Rs Aggarwal 2019 for Class 10 Math Chapter 19

Answer:

(i) 0
(ii) 1
(iii) 1
(iv) 0, 1
(v) 1

Answer:

When a coin is tossed once, the possible outcomes are H and T.
Total number of possible outcomes = 2
Favourable outcome = 1
∴ Probability of getting a tail = P (T) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{1}{2}$

Answer:

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4

(i) Let E be the event of getting exactly one head.

Then, the favourable outcomes are HT and TH.
Number of favourable outcomes = 2

∴ P(getting exactly 1 head) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{2}{4} = \frac{1}{2}

(ii) Let E be the event of getting at most one head.

Then, the favourable outcomes are HT, TH and TT.
Number of favourable outcomes = 3

∴ P (getting at most 1 head) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{3}{4}

(iii) Let E be the event of getting at least one head.

Then, the favourable outcomes are HT, TH and HH
Number of favourable outcomes = 3

∴ P (getting at least 1 head) =

\frac{Number of favourable outcomes}{Total number of possible outcomes} = \frac{3}{4}

Answer:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6

(i)
Let E be the event of getting an even number.
Then, the favourable outcomes are 2, 4 and 6.
Number of favourable outcomes = 3
∴ Probability of getting an even number = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{6} = \frac{1}{2}$
(ii)
Let E be the event of getting a number less than 5.
Then, the favourable outcomes are 1, 2, 3, 4
Number of favourable outcomes = 4
∴ Probability of getting a number less than 5 = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{4}{6} = \frac{2}{3}$

(iii)
Let E be the event of getting a number greater than 2.
Then, the favourable outcomes are 3, 4, 5 and 6.
Number of favourable outcomes = 4
∴ Probability of getting a number greater than 2 = P (E) = $\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{4}{6} = \frac{2}{3}$
(iv)
Let E be the event of getting a number between 3 and 6.

Then, the favourable outcomes are 4, 5

Number of favourable outcomes = 2
∴ Probability of getting a number between 3 and 6 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{2}{6} = \frac{1}{3}

(v)
Let E be the event of getting a number other than 3.

Then, the favourable outcomes are 1, 2, 4, 5 and 6.

Number of favourable outcomes = 5
∴ Probability of getting a number other than 3 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{5}{6}

(vi)
Let E be the event of getting the number 5.

Then, the favourable outcome is 5.

Number of favourable outcomes = 1
∴ Probability of getting the number 5 = P (E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{1}{6}

Answer:

Let E be the event of getting a consonant.

Out of 26 letters of English alphabets, there are 21 consonants.

∴ P(getting a consonant) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{21}{26}$

Thus, the probability of getting a consonant is $\frac{21}{26}$ .

Answer:

(i) The probability of getting A = P(A) = $\frac{number of favorable outcomes}{total number of possible outcomes} = \frac{3}{6} = \frac{1}{2} .$

(ii) The probability of getting D = P(D) = $\frac{number of favorable outcomes}{total number of possible outcomes} = \frac{1}{6} .$

Answer:

Total number of possible outcomes = 200

(i) Number of defective bulbs = 16
∴ P(getting a defective bulb) = $\frac{16}{200} = \frac{2}{25}$

(ii) Number of non-defective bulbs = 200 − 16 = 184
∴ P(getting a non-defective bulb) = $\frac{184}{200} = \frac{23}{25}$

Answer:

For any event E, P(E) + P(not E) = 1

Let probability of winning a game = P(E) = 0.7

∴ P(winning a game) + P(losing a game) = 1
⇒ P(losing a game) = 1 − 0.7
= 0.3

Thus, the probability of losing a game is 0.3.

Answer:

Total number of students = 35
Number of boys = 20
Number of girls = 15

(i) Let E₁ be the event that the chosen student is a boy.

∴ P(choosing a boy) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{20}{35} = \frac{4}{7}$

Thus, the probability that the chosen student is a boy is $\frac{4}{7}$ .

(ii) Let E₂ be the event that the chosen student is a girl.

∴ P(choosing a girl) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{35} = \frac{3}{7}$

Thus, the probability that the chosen student is a girl is $\frac{3}{7}$ .

Answer:

Total number of lottery tickets = 10 + 25 = 35
Number of prizes = 10

Let E be the event of getting a prize.

∴ P(getting a prize) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{10}{35} = \frac{2}{7}$

Thus, the probability of getting a prize is $\frac{2}{7}$ .

Answer:

Total number of tickets = 250
Kunal wins a prize if he gets a ticket that assures a prize.
Number of tickets on which prizes are assured = 5
∴ P (Kunal wins a prize) = $\frac{5}{250} = \frac{1}{50}$

Answer:

Total number of cards = 17

(i) Let E₁ be the event of choosing an odd number.

These numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17.

∴ P(getting an odd number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{9}{17}$

Thus, the probability that the card drawn bears an odd number is $\frac{9}{17}$ .

(i) Let E₂ be the event of choosing a number divisible by 5.

These numbers are 5, 10 and 15.
∴ P(getting a number divisible by 5) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{3}{17}$

Thus, the probability that the card drawn bears a number divisible by 5 is $\frac{3}{17}$ .

Answer:

Number of all possible outcomes = 8

Let E be the event of getting any factor of 8.

These numbers are 1, 2, 4 and 8.
∴ P(arrow will point at any factor of 8) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{4}{8} = \frac{1}{2}$

Thus, the probability that the arrow will point at any factor of 8 is $\frac{1}{2}$ .

Answer:

All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.

Number of all possible outcomes = 8

Let E be the event of having at least one boy.

Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.

Number of possible outcomes = 7
∴ P(Having at least one boy) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{7}{8}$

Thus, the probability of having at least one boy is $\frac{7}{8}$ .

Answer:

Total number of balls = 15

(i) Number of black balls = 2

∴  P(getting a black ball) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{15}$

Thus, the probability of getting a black ball is $\frac{2}{15}$ .

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

∴  P(getting a ball which is not green) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{11}{15}$

Thus, the probability of getting a ball which is not green is $\frac{11}{15}$ .

(iii) Number of balls which are either red or white = 4 + 5 = 9

∴  P(getting a ball which is red or white) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{9}{15} = \frac{3}{5}$

Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$ .

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

∴  P(getting a ball which is neither red nor green) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{15} = \frac{2}{5}$

Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$ .

Answer:

Total number of cards = 52

(i) Number of red kings = 2

∴ P(getting a red king) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a red king is $\frac{1}{26}$ .

(ii) Number of queens or jacks = 4 + 4 = 8

∴ P(getting a queen or a jack) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{52} = \frac{2}{13}$

Thus, the probability of getting a queen or a jack is $\frac{2}{13}$ .

Answer:

Favourable outcomes = 52 − 4kings − 4queens = 44

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{44}{52} = \frac{11}{13}$

Thus, the probability that the drawn card is neither a king nor a queen is $\frac{11}{13}$ .

Answer:

(i) Favourable outcomes = 2red kings + 2red queens + 2red jack = 6

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{6}{52} = \frac{3}{26}$

Thus, the probability of getting a red face card is $\frac{3}{26}$ .

(ii) Favourable outcomes = 2 (because there are 4 kings, 2 black and 2 red)

Total outcomes = 52

$P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes} = \frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a black king is $\frac{1}{26}$ .

Answer:

When two different dice are thrown, then total number of outcomes = 36.

(i) Let E₁ be the event of getting an even number on each die.

These numbers are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4) and (6, 6).

Number of Favourable outcomes = 9

∴ P(getting an even number on both dice) = $P (E_{1}) = \frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$

= $\frac{9}{36} = \frac{1}{4}$

Thus, the probability of getting an even number on both dice is $\frac{1}{4}$ .

(ii) Let E₂ be the event of getting the sum of the numbers appearing on the two dice is 5.

These numbers are (1, 4), (2, 3), (3, 2) and (4, 1)

Number of Favourable outcomes = 4

∴ P(getting the sum of the numbers appearing on the two dice is 5) = $P (E_{2}) = \frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$

= $\frac{4}{36} = \frac{1}{9}$

Thus, the probability of getting the sum of the numbers appearing on the two dice is 5 is $\frac{1}{9}$ .

Answer:

When two different dice are thrown, then total number of outcomes = 36.

Let E₁ be the event of getting the sum of the numbers on the two dice is 10.

These numbers are (4, 6), (5, 5) and (6, 4).

Number of Favourable outcomes = 3

∴ P(getting the sum of the numbers on the two dice is 10) = $P (E_{1}) = \frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$

= $\frac{3}{36} = \frac{1}{12}$

Thus, the probability of getting the sum of the numbers on the two dice is 10 is $\frac{1}{12}$ .

Answer:

Total number of possible outcomes is 36.
(i) The favorable outcomes for sum of numbers on dices less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1).
P(the sum of numbers appeared is less than 7)= $\frac{15}{36} = \frac{5}{12} .$

(ii) The favorable outcomes for product of numbers on dices less than 18 are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (5,1), (5,2), (5,3), (6,1), (6,2).
P(the product of numbers appeared is less than 18)= $\frac{26}{36} = \frac{13}{18} .$

Answer:

When two different dice are thrown, then total number of outcomes = 36.

Let E be the event of getting the product of numbers, as a perfect square.

These numbers are (1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5) and (6, 6).

Number of Favourable outcomes = 8

∴ P(getting the product of numbers, as a perfect square) = $P (E) = \frac{Number of outcomes favourable to E}{Number of all possible outcomes}$

= $\frac{8}{36} = \frac{2}{9}$

Thus, the probability of getting the product of numbers, as a perfect square is $\frac{2}{9}$ .

Answer:

Number of all possible outcomes = 36

Let E be the event of getting all those numbers whose product is 12.

These numbers are (2, 6), (3, 4), (4, 3) and (6, 2).

∴ P(getting all those numbers whose product is 12) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{4}{36} = \frac{1}{9}$

Thus, the probability of getting all those numbers whose product is 12 is $\frac{1}{9}$ .

Answer:

All possible outcomes are 5, 6, 7, 8...................50.

Number of all possible outcomes = 46

(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.
Let E₁ be the event of getting a prime number less than 10.
Then, number of favourable outcomes = 2
∴ P (getting a prime number less than 10) =

\frac{2}{46} = \frac{1}{23}

(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.
   Let E₂ be the event of getting a perfect square.
Then, number of favourable outcomes = 5
∴ P (getting a perfect square) =

\frac{5}{46}

Answer:

The possible outcomes are 1,2, 3,4, 5................12.
Number of all possible outcomes = 12

(i) Let E₁ be the event that the pointer rests on 6.
Then, number of favourable outcomes = 1
∴ P (arrow pointing at 6) = P( E₁) = $\frac{1}{12}$

(ii) Out of the given numbers, the even numbers are

2, 4, 6, 8,10 and 12

Let E₂ be the event of getting an even number.
Then, number of favourable outcomes = 6
∴ P (arrow pointing at an even number) = P( E₂) =

\frac{6}{12} = \frac{1}{2}

(iii) Out of the given numbers, the prime numbers are 2, 3, 5, 7 and 11.
   Let E₃ be the event of the arrow pointing at a prime number.
  Then, number of favourable outcomes = 5
    ∴ P (arrow pointing at a prime number) = P( E₃)   =

\frac{5}{12}

(iv) Out of the given numbers, the numbers that are multiples of 5 are 5 and 10 only.
Let E₄ be the event of the arrow pointing at a multiple of 5.

Then, number of favourable outcomes = 2
∴ P(arrow pointing at a number that is a multiple of 5) = P( E₄) =

\frac{2}{12} = \frac{1}{6}

Answer:

Total number of pens = 132 + 12 = 144

Number of good pens = 132

Let E be the event of getting a good pen.

∴ P(getting a good pen) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{132}{144} = \frac{11}{12}$

Thus, the probability of getting a good pen is $\frac{11}{12}$ .

Answer:

Total number of pens = 144
Number of defective pens = 20
Number of good pens = 144 − 20 = 124

(i) Let E₁ be the event of getting a good pen.

∴ P(buying a pen) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{124}{144} = \frac{31}{36}$

Thus, the probability that Tanvy will buy a pen is $\frac{31}{36}$ .

(ii) Let E₂ be the event of getting a defective pen.

∴ P(not buying a pen) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{20}{144} = \frac{5}{36}$

Thus, the probability that Tanvy will not buy a pen is $\frac{5}{36}$ .

Answer:

Total number of discs = 90

(i) Let E₁ be the event of having a two-digit number.

Number of discs bearing two-digit number = 90 − 9 = 81

∴ P(getting a two-digit number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{81}{90} = \frac{9}{10}$

Thus, the probability that the disc bears a two-digit number is $\frac{9}{10}$ .

(ii) Let E₂ be the event of getting a perfect square number.

Discs bearing perfect square numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Number of discs bearing a perfect square number = 9

∴ P(getting a perfect square number) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{9}{90} = \frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$ .

(iii) Let E₃ be the event of getting a number divisible by 5.

Discs bearing numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90.

Number of discs bearing a number divisible by 5 = 18

∴ P(getting a number divisible by 5) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{18}{90} = \frac{1}{5}$

Thus, the probability that the disc bears a number divisible by 5 is $\frac{1}{5}$ .

Answer:

(i) Number of all possible outcomes = 20.
Number of defective bulbs = 4.
Number of non-defective bulbs = 20 − 4 = 16.

Let E₁ be the event of getting a defective bulb.

∴ P(getting a defective bulb) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{4}{20} = \frac{1}{5}$

Thus, the probability that the bulb is defective is $\frac{1}{5}$ .

(ii) After removing 1 non-defective bulb, we have number of remaining bulbs = 19.
Out of these, number of non-defective bulbs = 16 − 1 = 15.

Let E₂ be the event of getting a non-defective bulb.

∴ P(getting a non-defective bulb) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{19}$

Thus, the probability that the bulb is non-defective is $\frac{15}{19}$ .

Answer:

Suppose there are x candies in the bag.
Then, number of orange candies in the bag = 0.
And, number of lemon candies in the bag = x.

(i) Let E₁ be the event of getting an orange-flavoured candy.

∴ P(getting an orange-flavoured candy) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{0}{x} = 0$

Thus, the probability that Hema takes out an orange-flavoured candy is 0.

(ii) Let E₂ be the event of getting a lemon-flavoured candy.

∴ P(getting a lemon-flavoured candy) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{x}{x} = 1$

Thus, the probability that Hema takes out a lemon-flavoured candy is 1.

Answer:

Total number of students = 40.
Number of boys = 15.
Number of girls = 25.

(i) Let E₁ be the event of getting a girl's name on the card.

∴ P(selecting the name of a girl) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{25}{40} = \frac{5}{8}$

Thus, the probability that the name written on the card is the name of a girl is $\frac{5}{8}$ .

(ii) Let E₂ be the event of getting a boy's name on the card.

∴ P(selecting the name of a boy) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{15}{40} = \frac{3}{8}$

Thus, the probability that the name written on the card is the name of a boy is $\frac{3}{8}$ .

Answer:

Total number of all possible outcomes= 52
(i) Total number of aces = 4
∴ P( getting an ace) = $\frac{4}{52} = \frac{1}{13}$

(ii) Number of 4 of spades = 1
∴ P(getting a 4 of spade) = $\frac{1}{52}$

(iii) Number of 9 of a black suit = 2
∴ P(getting a 9 of a black suit) = $\frac{2}{52} = \frac{1}{26}$

(iv) Number of red kings = 2
∴ P(getting a red king) = $\frac{2}{52} = \frac{1}{26}$

Answer:

Total number of all possible outcomes= 52
(i) Total number of queens = 4
∴ P(getting a queen) = $\frac{4}{52} = \frac{1}{13}$

(ii) Number of diamond suits = 13
∴ P(getting a diamond) = $\frac{13}{52} = \frac{1}{4}$

(iii) Total number of kings = 4
Total number of aces = 4
Let E be the event of getting a king or an ace card.
Then, the favourable outcomes = 4 + 4 = 8
∴ P( getting a king or an ace) = P (E) = $\frac{8}{52} = \frac{2}{13}$

(iv) Number of red aces = 2
∴ P( getting a red ace) = $\frac{2}{52} = \frac{1}{26}$

Answer:

Total number of outcomes = 52

(i) Let E₁ be the event of getting a king of red suit.

Number of favourable outcomes = 2

∴ P(getting a king of red suit) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a king of red suit is $\frac{1}{26}$ .

(ii) Let E₂ be the event of getting a face card.

Number of favourable outcomes = 12

∴ P(getting a face card) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{12}{52} = \frac{3}{13}$

Thus, the probability of getting a face card is $\frac{3}{13}$ .

(iii) Let E₃ be the event of getting a red face card.

Number of favourable outcomes = 6

∴ P(getting a red face card) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{6}{52} = \frac{3}{26}$

Thus, the probability of getting a red face card is $\frac{3}{26}$ .

(iv) Let E₄ be the event of getting a queen of black suit.

Number of favourable outcomes = 2

∴ P(getting a queen of black suit) = P(E₄) = $\frac{Number of outcomes favourable to E_{4}}{Number of all possible outcomes}$
= $\frac{2}{52} = \frac{1}{26}$

Thus, the probability of getting a queen of black suit is $\frac{1}{26}$ .

(v) Let E₅ be the event of getting a jack of hearts.

Number of favourable outcomes = 1

∴ P(getting a jack of hearts) = P(E₅) = $\frac{Number of outcomes favourable to E_{5}}{Number of all possible outcomes}$
= $\frac{1}{52}$

Thus, the probability of getting a jack of hearts is $\frac{1}{52}$ .

(vi) Let E₆ be the event of getting a spade.

Number of favourable outcomes = 13

∴ P(getting a spade) = P(E₆) = $\frac{Number of outcomes favourable to E_{6}}{Number of all possible outcomes}$
= $\frac{13}{52} = \frac{1}{4}$

Thus, the probability of getting a spade is $\frac{1}{4}$ .

Answer:

Total number of all possible outcomes= 52
(i) Number of spade cards = 13
Number of aces = 4 (including 1 of spade)

Therefore, number of spade cards and aces = (13 + 4 − 1) = 16

∴ P( getting a spade or an ace card) =

\frac{16}{52} = \frac{4}{13}

(ii) Number of red kings = 2
∴ P( getting a red king) =

\frac{2}{52} = \frac{1}{26}

(iii) Total number of kings = 4
Total number of queens = 4
Let E be the event of getting either a king or a queen.
Then, the favourable outcomes = 4 + 4 = 8
∴ P( getting a king or a queen) = P (E) =

\frac{8}{52} = \frac{2}{13}

(iv) Let E be the event of getting either a king or a queen. Then, ( not E) is the event that drawn card is neither a king nor a

queen.

Then, P(getting a king or a queen ) =

\frac{2}{13}

Now, P (E) + P (not E) = 1
∴ P(getting neither a king nor a queen ) =

1 - (\frac{2}{13}) = \frac{11}{13}

Answer:

Total number of possible outcomes is 36.
(i) The favorable outcomes are (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6).
P(the sum is even)= $\frac{18}{36} = \frac{1}{2} .$

(ii) The favorable outcomes are (1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).
P(the product is even)= $\frac{27}{36} = \frac{3}{4} .$

Answer:

Total number of possible outcomes is 36.
(i) The favorable outcomes for sum of numbers on dices less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1).
P(the sum of numbers appeared is less than 7)= $\frac{15}{36} = \frac{5}{12} .$

(ii) The favorable outcomes for product of numbers on dices less than 16 are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2).
P(the product of numbers appeared is less than 16)= $\frac{25}{36} .$

(iii) The favorable outcomes are (1,1), (3,3), (5,5).
P(the doublet of odd numbers)= $\frac{3}{36} = \frac{1}{12} .$

Answer:

Total number of cards in a deck is 52.
The number of cards left after loosing the King, the Jack and the 10 of spade = 52 $-$ 3 = 49.
(i) Red card left after loosing King, the Jack and the 10 of spade will be 13 hearts + 13 diamond = 26 cards
P(red card)= $\frac{26}{49} .$
(ii) Black jack left after loosing King, the Jack and the 10 of spade will be one only of club.
P(black jack)= $\frac{1}{49} .$
(iii) Red king left after loosing King, the Jack and the 10 of spade will be red of diamond and red of hearts i.e two cards.
P(red king)= $\frac{2}{49} .$
(iv) 10 of hearts will be one only after loosing King, the Jack and the 10 of spade
P(10 of hearts)= $\frac{1}{49} .$

Answer:

Total number of possible outcomes for Peter = 36.
Possible outcomes for Peter to get product 25 is (5,5).
Total number of possible outcomes for Rina = 6.
Possible outcomes for Rina to get the number whose square is 25 is 5.

Now, P(Peter will get 25)= $\frac{1}{36} .$
And, P(Rina will get 25)= $\frac{1}{6} .$
Since $\frac{1}{36} < \frac{1}{6}$ , So Rina has better chances of getting 25.

Answer:

Total number of outcomes = 25

(i) Let E₁ be the event of getting a card divisible by 2 or 3.

Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.
Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favourable outcomes = 16

∴ P(getting a card divisible by 2 or 3) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{16}{25}$

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\frac{16}{25}$ .

(ii) Let E₂ be the event of getting a prime number.

Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favourable outcomes = 9

∴ P(getting a prime number) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{9}{25}$

Thus, the probability that the number on the drawn card is a prime number is $\frac{9}{25}$ .

Answer:

Given numbers 3, 5, 7, 9, .... , 35, 37 form an AP with a = 3 and d = 2.
Let T_n= 37. Then,
3 + (n − 1)2 = 37
⇒ 3 + 2n − 2 = 37
⇒ 2n = 36
⇒ n = 18

Thus, total number of outcomes = 18.

Let E be the event of getting a prime number.

Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.

Number of favourable outcomes = 11.

∴ P(getting a prime number) = P(E) = $\frac{Number of outcomes favourable to E}{Number of all possible outcomes}$
= $\frac{11}{18}$

Thus, the probability that the number on the card is a prime number is $\frac{11}{18}$ .

Answer:

Total number of outcomes = 30.

(i) Let E₁ be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favourable outcomes = 30 − 10 = 20

∴ P(getting a number not divisible by 3) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{20}{30} = \frac{2}{3}$

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$ .

(ii) Let E₂ be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 6

∴ P(getting a prime number greater than 7) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{6}{30} = \frac{1}{5}$

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$ .

(iii) Let E₃ be the event of getting a number which is not a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16 and 25.

Number of favourable outcomes = 30 − 5 = 25

∴ P(getting non-perfect square number) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{25}{30} = \frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$ .

Answer:

Given number 1, 3, 5, .... , 35 form an AP with a = 1 and d = 2.
Let T_n = 35. Then,
1 + (n − 1)2 = 35
⇒ 1 + 2n − 2 = 35
⇒ 2n = 36
⇒ n = 18

Thus, total number of outcomes = 18.

(i) Let E₁ be the event of getting a prime number less than 15.

Out of these numbers, prime numbers less than 15 are 3, 5, 7, 11 and 13.

Number of favourable outcomes = 5.

∴ P(getting a prime number less than 15) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{5}{18}$

Thus, the probability of getting a card bearing a prime number less than 15 is $\frac{5}{18}$ .

(ii) Let E₂ be the event of getting a number divisible by 3 and 5.

Out of these numbers, number divisible by 3 and 5 means number divisible by 15 is 15.

Number of favourable outcomes = 1.

∴ P(getting a number divisible by 3 and 5) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{1}{18}$

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is $\frac{1}{18}$ .

Answer:

Given number 6, 7, 8, .... , 70 form an AP with a = 6 and d = 1.
Let T_n = 70. Then,
6 + (n − 1)1 = 70
⇒ 6 + n  − 1 = 70
⇒ n = 65

Thus, total number of outcomes = 65.

(i) Let E₁ be the event of getting a one-digit number.

Out of these numbers, one-digit numbers are 6, 7, 8 and 9.

Number of favourable outcomes = 4.

∴ P(getting a one-digit number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{4}{65}$

Thus, the probability that the card bears a one-digit number is $\frac{4}{65}$ .

(ii) Let E₂ be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ... , 70.
Given number 10, 15, 20, .... , 70 form an AP with a = 10 and d = 5.
Let T_n = 70. Then,
10 + (n − 1)5 = 70
⇒ 10 + 5n  − 5 = 70
⇒ 5n = 65
⇒ n = 13

Thus, number of favourable outcomes = 13.

∴ P(getting a number divisible by 5) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{13}{65} = \frac{1}{5}$

Thus, the probability that the card bears a number divisible by 5 is $\frac{1}{5}$ .

(iii) Let E₃ be the event of getting an odd number less than 30.

Out of these numbers, odd numbers less than 30 are 7, 9, 11, ... , 29.
Given number 7, 9, 11, .... , 29 form an AP with a = 7 and d = 2.
Let T_n = 29. Then,
7 + (n − 1)2 = 29
⇒ 7 + 2n  − 2 = 29
⇒ 2n = 24
⇒ n = 12

Thus, number of favourable outcomes = 12.

∴ P(getting an odd number less than 30) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{12}{65}$

Thus, the probability that the card bears an odd number less than 30 is $\frac{12}{65}$ .

(iv) Let E₄ be the event of getting a composite number between 50 and 70.

Out of these numbers, composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Number of favourable outcomes = 15.

∴ P(getting a composite number between 50 and 70) = P(E₄) = $\frac{Number of outcomes favourable to E_{4}}{Number of all possible outcomes}$
= $\frac{15}{65} = \frac{3}{13}$

Thus, the probability that the card bears a composite number between 50 and 70 is $\frac{3}{13}$ .

Answer:

Given number 1, 3, 5, ..., 101 form an AP with a = 1 and d = 2.
Let T_n = 101. Then,
1 + (n − 1)2 = 101
⇒ 1 + 2n − 2 = 101
⇒ 2n = 102
⇒ n = 51

Thus, total number of outcomes = 51.

(i) Let E₁ be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, ... , 17.
Given number 1, 3, 5, .... , 17 form an AP with a = 1 and d = 2.
Let T_n = 17. Then,
1 + (n − 1)2 = 17
⇒ 1 + 2n − 2 = 17
⇒ 2n = 18
⇒ n = 9

Thus, number of favourable outcomes = 9.

∴ P(getting a number less than 19) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{9}{51} = \frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$ .

(ii) Let E₂ be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, number of favourable outcomes = 7.

∴ P(getting a prime number less than 20) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$ .

Answer:

All possible outcomes are 2, 3, 4, 5................101.
Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8...................100
Let E₁ be the event of getting an even number.
Then, number of favourable outcomes = 50    $[T_{n} = 100 \Rightarrow 2 + (n - 1) \times 2 = 100, \Rightarrow n = 50]$
∴ P (getting an even number) = $\frac{50}{100} = \frac{1}{2}$

(ii) Out of these, the numbers that are less than 16 = 2,3,4,5,..................15.
Let E₂ be the event of getting a number less than 16.
Then, number of favourable outcomes = 14
    ∴ P (getting a number less than 16) = $\frac{14}{100} = \frac{7}{50}$

(iii) Out of these, the numbers that are perfect squares = 4, 9,16,25, 36, 49, 64, 81 and 100
  Let E₃ be the event of getting a number that is a perfect square.

Then, number of favourable outcomes = 9
∴ P (getting a number that is a perfect square) =

\frac{9}{100}

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Let E₄ be the event of getting a prime number less than 40.
Then, number of favourable outcomes = 12
∴ P (getting a prime number less than 40) =

\frac{12}{100} = \frac{3}{25}

Answer:

(i)
Total number of outcomes = 80.

Let E₁ be the event of getting a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16, 25, 36, 49 and 64.

Thus, number of favourable outcomes = 8.

∴ P(getting a perfect square number) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{8}{80} = \frac{1}{10}$

Thus, the probability that the disc bears a perfect square number is $\frac{1}{10}$ .

(ii)
Total possible outcomes = 90.
(a) Two-digit numbers are 10,11,12,13...90.
P(getting a two-digit number)= $\frac{81}{90} = \frac{9}{10} .$

(b) Numbers divisible by 5 are 5,10,15,20...90.
P(number divisible by 5)= $\frac{18}{90} = \frac{1}{5} .$

Answer:

Number of 50 p coins = 100.
Number of ₹1 coins = 70.
Number of ₹2 coins = 50.
Number of ₹5 coins = 30.

Thus, total number of outcomes = 250.

(i) Let E₁ be the event of getting a ₹1 coin.

Number of favourable outcomes = 70.

∴ P(getting a ₹1 coin) = P(E₁) = $\frac{Number of outcomes favourable to E_{1}}{Number of all possible outcomes}$
= $\frac{70}{250} = \frac{7}{25}$

Thus, the probability that the coin will be a ₹1 coin is $\frac{7}{25}$ .

(ii) Let E₂ be the event of not getting a ₹5 coin.

Number of favourable outcomes = 250 − 30 = 220

∴ P(not getting a ₹5 coin) = P(E₂) = $\frac{Number of outcomes favourable to E_{2}}{Number of all possible outcomes}$
= $\frac{220}{250} = \frac{22}{25}$

Thus, the probability that the coin will not be a ₹5 coin is $\frac{22}{25}$ .

(iii) Let E₃ be the event of getting a 50 p or a ₹2 coin.

Number of favourable outcomes = 100 + 50 = 150

∴ P(getting a 50 p or a ₹2 coin) = P(E₃) = $\frac{Number of outcomes favourable to E_{3}}{Number of all possible outcomes}$
= $\frac{150}{250} = \frac{3}{5}$

Thus, the probability that the coin will be a 50 p or a ₹2 coin is $\frac{3}{5}$ .

Answer:

It is given that,

P(getting a red ball) = $\frac{1}{4}$ and P(getting a blue ball) = $\frac{1}{3}$

Let P(getting an orange ball) be x.

Since, there are only 3 types of balls in the jar, the sum of probabilities of all the three balls must be 1.

$∴ \frac{1}{4} + \frac{1}{3} + x = 1 \Rightarrow x = 1 - \frac{1}{4} - \frac{1}{3} \Rightarrow x = \frac{12 - 3 - 4}{12} \Rightarrow x = \frac{5}{12}$ \
∴ P(getting an orange ball) = $\frac{5}{12}$

Let the total number of balls in the jar be n.

∴ P(getting an orange ball) = $\frac{10}{n}$
$\Rightarrow \frac{10}{n} = \frac{5}{12} \Rightarrow n = 24$

Thus, the total number of balls in the jar is 24.

Answer:

Total number of balls = 18.
Number of red balls = x.

(i) Number of balls which are not red = 18 − x

∴ P(getting a ball which is not red) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{18 - x}{18}$

Thus, the probability of drawing a ball which is not red is $\frac{18 - x}{18}$ .

(ii) Now, total number of balls = 18 + 2 = 20.
Number of red balls now = x + 2.

P(getting a red ball now) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{x + 2}{20}$

and P(getting a red ball in first case) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{x}{18}$

Since, it is given that probability of drawing a red ball now will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case.

Thus, $\frac{x + 2}{20} = \frac{9}{8} \times \frac{x}{18}$
$\Rightarrow 144 (x + 2) = 180 x \Rightarrow 144 x + 288 = 180 x \Rightarrow 36 x = 288 \Rightarrow x = \frac{288}{36} = 8$

Thus, the value of x is 8.

Answer:

Total number of marbles = 24.
Let the number of blue marbles be x.
Then, the number of green marbles = 24 − x

∴ P(getting a green marble) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{24 - x}{24}$

But, P(getting a green marble) = $\frac{2}{3}$ (given)

$∴ \frac{24 - x}{24} = \frac{2}{3} \Rightarrow 3 (24 - x) = 48 \Rightarrow 72 - 3 x = 48 \Rightarrow 3 x = 72 - 48 \Rightarrow 3 x = 24 \Rightarrow x = 8$

Thus, the number of blue marbles in the jar is 8.

Answer:

Total number of marbles = 54.

It is given that, P(getting a blue marble) = $\frac{1}{3}$ and P(getting a green marble) = $\frac{4}{9}$

Let P(getting a white marble) be x.

Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.

$∴ \frac{1}{3} + \frac{4}{9} + x = 1 \Rightarrow \frac{3 + 4}{9} + x = 1 \Rightarrow x = 1 - \frac{7}{9} \Rightarrow x = \frac{2}{9}$
∴ P(getting a white marble) = $\frac{2}{9}$ ...(1)

Let the number of white marbles be n.

Then, P(getting a white marble) = $\frac{n}{54}$ ... (2)

From (1) and (2),
$\frac{n}{54} = \frac{2}{9} \Rightarrow n = \frac{2 \times 54}{9} \Rightarrow n = 12$

Thus, there are 12 white marbles in the jar.

Answer:

Total number of shirts = 100.
Number of good shirts = 88.
Number of shirts with minor defects = 8.
Number of shirts with major defects = 100 − 88 − 8 = 4.

(i) P(the drawn shirt is acceptable to Rohit) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{88}{100} = \frac{22}{25}$

Thus, the probability that the drawn shirt is acceptable to Rohit is $\frac{22}{25}$ .

(ii) P(the drawn shirt is acceptable to Kamal) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{88 + 8}{100} = \frac{96}{100} = \frac{24}{25}$

Thus, the probability that the drawn shirt is acceptable to Kamal is $\frac{24}{25}$ .

Answer:

Total number of persons = 12.
Number of persons who are extremely patient = 3.
Number of persons who are extremely honest = 6.
Number of persons who are extremely kind = 12 − 3 − 6 = 3.

(i) P(selecting a person who is extremely patient) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{12} = \frac{1}{4}$

Thus, the probability of selecting a person who is extremely patient is $\frac{1}{4}$ .

(ii) P(selecting a person who is extremely kind or honest) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6 + 3}{12} = \frac{9}{12} = \frac{3}{4}$

Thus, the probability of selecting a person who is extremely kind or honest is $\frac{3}{4}$ .

From the three given values, we prefer honesty more.

Answer:

Total number of outcomes = 36.

(i) Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

Number of such cases = 11.

Number of cases where 5 will not come up either time = 36 − 11 = 25.

∴ P(5 will not come up either time) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{25}{36}$

Thus, the probability that 5 will not come up either time is $\frac{25}{36}$ .

(ii) Cases where 5 comes up on exactly one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).

Number of such cases = 10.

∴ P(5 will come up exactly one time) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{10}{36} = \frac{5}{18}$

Thus, the probability that 5 will come up exactly one time is $\frac{5}{18}$ .

(iii) Cases where 5 comes up on exactly two times is (5, 5).

Number of such cases = 1.

∴ P(5 will come up both the times) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{36}$

Thus, the probability that 5 will come up both the times is $\frac{1}{36}$ .

Answer:

Number of possible outcomes = 36

Let E be the event of getting two numbers whose product is a perfect square.
Then, the favourable outcomes are (1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Number of favourable outcomes = 8.

∴ P(getting numbers whose product is a perfect square) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{36} = \frac{2}{9}$

Thus, the probability of getting such numbers on two dice whose product is a perfect square is $\frac{2}{9}$ .

Answer:

Total numbers of letters in the given word ASSOCIATION = 11
(i) Number of vowels ( A, O, I, A, I, O) in the given word = 6
∴ P (getting a vowel) = $\frac{6}{11}$

(ii) Number of consonants in the given word ( S, S, C, T, N) = 5
∴ P (getting a consonant) = $\frac{5}{11}$

(iii) Number of S in the given word = 2
∴ P (getting an S) = $\frac{2}{11}$

Answer:

Total number of cards = 5.

(a) Number of queens = 1.

∴ P(getting a queen) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{5}$

Thus, the probability that the drawn card is the queen is $\frac{1}{5}$ .

(b) When the queen is put aside, number of remaining cards = 4.

(i) Number of aces = 1.

∴ P(getting an ace) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{4}$

Thus, the probability that the drawn card is an ace is $\frac{1}{4}$ .

(ii) Number of queens = 0.

∴ P(getting a queen now) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{0}{4} = 0$

Thus, the probability that the drawn card is a queen is 0.

Answer:

Total number of all possible outcomes= 52
There are 26 red cards (including 2 queens) and apart from these, there are 2 more queens.
Number of cards, each one of which is either a red card or a queen = 26 + 2 = 28
Let E be the event that the card drawn is neither a red card nor a queen.
Then, the number of favourable outcomes = (52 − 28) = 24
∴ P( getting neither a red card nor a queen) = P (E) = $\frac{24}{52} = \frac{6}{13}$

Answer:

An ordinary year has 365 days consisting of 52 weeks and 1 day.

This day can be any day of the week.

∴ P(of this day to be Monday) = $\frac{1}{7}$

Thus, the probability that an ordinary year has 53 Mondays is $\frac{1}{7}$ .

Answer:

There are 6 red face cards. These are removed.

Thus, remaining number of cards = 52 − 6 = 46.

(i) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{20}{46} = \frac{10}{23}$

Thus, the probability that the drawn card is a red card is $\frac{10}{23}$ .

(ii) Number of face cards now = 12 − 6 = 6.

∴ P(getting a face card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{46} = \frac{3}{23}$

Thus, the probability that the drawn card is a face card is $\frac{3}{23}$ .

(iii) Number of card of clubs = 12.

∴ P(getting a card of clubs) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{12}{46} = \frac{6}{23}$

Thus, the probability that the drawn card is a card of clubs is $\frac{6}{23}$ .

Answer:

There are 4 kings, 4 queens and 4 aces. These are removed.

Thus, remaining number of cards = 52 − 4 − 4 − 4 = 40.

(i) Number of black face cards now = 2 (only black jacks).

∴ P(getting a black face card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{40} = \frac{1}{20}$

Thus, the probability that the drawn card is a black face card is $\frac{1}{20}$ .

(ii) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{20}{40} = \frac{1}{2}$

Thus, the probability that the drawn card is a red card is $\frac{1}{2}$ .

Answer:

When a coin is tossed three times, all possible outcomes are
HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.

Number of total outcomes = 8.

(i) Outcome with three heads is HHH.

Number of outcomes with three heads = 1.

∴ P(getting three heads) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{8}$

Thus, the probability of getting three heads is $\frac{1}{8}$ .

(ii) Outcomes with atleast two tails are TTH, THT, HTT and TTT.

Number of outcomes with atleast two tails = 4.

∴ P(getting at least two tails) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{4}{8} = \frac{1}{2}$

Thus, the probability of getting at least two tails is $\frac{1}{2}$ .

Answer:

A leap year has 366 days with 52 weeks and 2 days.

Now, 52 weeks conatins 52 sundays.

The remaining two days can be:
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday

Out of these 7 cases, there are two cases favouring it to be Sunday.

∴ P(a leap year having 53 Sundays) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{2}{7}$

Thus, the probability that a leap year selected at random will contain 53 Sundays is $\frac{2}{7}$ .

Answer:

Total number of apples = 900.
P(a rotten apple) = 0.18
$\Rightarrow \frac{number of rotten apples}{total number of apples} = 0.18 \Rightarrow \frac{number of rotten apples}{900} = 0.18 \Rightarrow The number of rotten apples = 0.18 \times 900 = 162 apples .$

Answer:

The number of white balls in a bag = 15.
Let the number of black balls in that bag be x.
Then, the total number of balls in bag = 15+x.

Now, P(black ball)= $\frac{x}{15 + x}$ and P(white ball)= $\frac{15}{15 + x}$ .
According to question, P(black ball) = 3 P(white ball)
$\Rightarrow \frac{x}{15 + x} = 3 \times \frac{15}{15 + x} \Rightarrow \frac{x}{15 + x} = \frac{45}{15 + x} \Rightarrow x = 45 .$
Hence, number of black balls in bag = x = 45.

Answer:

Favorable outcomes for sum of numbers on dice less than 3 or more than 11 are (1,1), (6,6).
Required probability = P(sum is less than 3 or more than 11) = $\frac{2}{36} = \frac{1}{18} .$

Answer:

Since, number of elements in the set of favourable cases is less than or equal to the number of elements in the set of whole number of cases,
their ratio always end up being 1 or less than 1.
Also, their ratio can never be negative.

Thus, probability of an event always lies between 0 and 1.
i.e. 0 ≤ P(E) ≤ 1

Hence, the correct answer is option (c).

Answer:

P(occurence of an event) = p
P(non-occurence of this event) = 1 − p

Hence, the correct answer is option (b).

Answer:

(b) 0

The probability of an impossible event is 0.

Answer:

(c) 1

The probability of a sure event is 1.

Answer:

The probability of an event cannot be greater than 1.
Thus, the probability of an event cannot be 1.5.

Hence, the correct answer is option (a).

Answer:

Total number of outcomes = 30.

Prime numbers in 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 10.

∴ P(getting a prime number) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{10}{30} = \frac{1}{3}$

Thus, the probability that the selected number is a prime number is $\frac{1}{3}$ .

Hence, the correct answer is option (c).

Answer:

Total number of outcomes = 15.

Out of the given numbers, multiples of 4 are 4, 8 and 12.

Numbers of favourable outcomes = 3.

∴ P(getting a multiple of 4) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{15} = \frac{1}{5}$

Thus, the probability that a number selected is a multiple of 4 is $\frac{1}{5}$ .

Hence, the correct answer is option (c).

Answer:

Total number of outcomes = 45.

Out of the given numbers, perfect square numbers are 9, 16, 25, 36 and 49.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a perfect square) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{45} = \frac{1}{9}$

Thus, the probability that the drawn card has a number which is a perfect square is $\frac{1}{9}$ .

Hence, the correct answer is option (d).

Answer:

Total number of discs = 90.

Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.

Numbers of favourable outcomes = 8.

∴ P(getting a prime number which is less than 23) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{8}{90} = \frac{4}{45}$

Thus, the probability that the disc bears prime number less than 23 is $\frac{4}{45}$ .

Disclaimer: There is a misprinting in the question. If they ask for the probability of all prime numbers less than 90, then we get $\frac{4}{15}$ .

Hence, the correct answer is option (c).

Answer:

Total number of cards = 10.

Out of the given numbers, prime numbers are 2, 3, 5, 7 and 11.

Numbers of favourable outcomes = 5.

∴ P(getting a prime number) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{10} = \frac{1}{2}$

Thus, the probability of getting a card with a prime number is $\frac{1}{2}$ .

Hence, the correct answer is option (a).

Answer:

Total number of tickets = 40.

Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a multiple of 7) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{5}{40} = \frac{1}{8}$

Thus, the probability that the selected ticket has a number, which is a multiple of 7, is $\frac{1}{8}$ .

Hence, the correct answer is option (b).

Answer:

(d) $\frac{7}{6}$

Explanation:
Probability of an event can't be more than 1.

Answer:

(c) 0.6

Explanation:
Let E be the event of winning a game.
Then, ( not E) is the event of not winning the game or of losing the game.
Then, P(E) = 0.4
Now, P(E) + P(not E) = 1
⇒ 0.4 + P(not E) = 1
⇒ P(not E) = 1− 0.4 = 0.6
∴ P(losing the game) = P(not E) = 0.6

Answer:

(d) 0

If an event cannot occur, its probability is 0.

Answer:

(b) $\frac{1}{5}$

Explanation:
Total number of tickets = 20
Out of the given ticket numbers, multiples of 5 are 5, 10, 15 and 20.
Number of favourable outcomes = 4
∴ P(getting a multiple of 5) = $\frac{4}{20} = \frac{1}{5}$

Answer:

(c) $\frac{12}{25}$

Explanation:
Total number of tickets = 25
Let E be the event of getting a multiple of 3 or 5.
Then,
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.
Multiples of 5 are 5, 10, 15, 20 and 25.

Number of favourable outcomes = ( 8 + 5 − 1) = 12
∴ P (getting a multiple of 3 or 5 ) = P (E) = $\frac{12}{25}$

Answer:

(d) $\frac{2}{5}$

Explanation:
All possible outcomes are 6, 7, 8................15.
Number of all possible outcomes = 10

Out of these, the numbers that are less than 10 are 6, 7, 8 and 9.
Number of favourable outcomes = 4
∴ P(getting a number that is less than 10) =

\frac{4}{10} = \frac{2}{5}

Answer:

Total number of outcomes = 6.

Out of the given numbers, even numbers are 2, 4 and 6.

Numbers of favourable outcomes = 3.

∴ P(getting an even number) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{3}{6} = \frac{1}{2}$

Thus, the probability of getting an even number is $\frac{1}{2}$ .

Hence, the correct answer is option (a).

Answer:

Total number of outcomes = 6.

Out of the given numbers, numbers greater than 2 are 3, 4, 5 and 6.

Numbers of favourable outcomes = 4.

∴ P(getting a number greater than 2) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{4}{6} = \frac{2}{3}$

Thus, the probability of getting a number greater than 2 is $\frac{2}{3}$ .

Hence, the correct answer is option (d).

Answer:

Total number of outcomes = 6.

Out of the given numbers, odd number greater than 3 is 5.

Numbers of favourable outcomes = 1.

∴ P(getting an odd number greater than 3) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{6}$

Thus, the probability of getting an odd number greater than 3 is $\frac{1}{6}$ .

Hence, the correct answer is option (b).

Answer:

(c) $\frac{1}{2}$

Explanation:
In a single throw of a die, the possible outcomes are:
1, 2, 3, 4, 5, 6
Total number of possible outcomes = 6
Let E be the event of getting a prime number.
Then, the favourable outcomes are 2, 3 and 5.
Number of favourable outcomes = 3
∴ Probability of getting a prime number = P (E) = $\frac{Number of favorable outcomes}{Total number of possible outcomes} = \frac{3}{6} = \frac{1}{2}$

Answer:

Total number of outcomes = 36.

Getting the same number on both the dice means (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Numbers of favourable outcomes = 6.

∴ P(getting the same number on both dice) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{6}{36} = \frac{1}{6}$

Thus, the probability of getting the same number on both dice is $\frac{1}{6}$ .

Hence, the correct answer is option (c).

Answer:

All possible outcomes are HH, HT, TH and TT.

Total number of outcomes = 4.

Getting 2 heads means getting HH.

Numbers of favourable outcomes = 1.

∴ P(getting 2 heads) = $\frac{Number of favourable outcomes}{Number of all possible outcomes}$
= $\frac{1}{4}$

Thus, the probability of getting 2 heads is $\frac{1}{4}$ .

Hence, the correct answer is option (d).

Answer:

(2) $\frac{1}{6}$

Explanation:
When two dice are thrown simultaneously, all possible outcomes are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Number of all possible outcomes = 36

Let E be the event of getting a doublet.
Then the favourable outcomes are:
(1,1), (2,2) , (3,3) (4,4), (5,5), (6,6)
Number of favourable outcomes = 6
∴ P(getting a doublet) = P ( E) = $\frac{6}{36} = \frac{1}{6}$

Answer:

(d) $\frac{3}{4}$

Explanation:
When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.
Total number of possible outcomes = 4

Let E be the event of getting at most one head.

Then, the favourable outcomes are HT, TH and TT.
Number of favourable outcomes = 3

∴ P(getting at most 1head) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{4}

Answer:

(c) $\frac{3}{8}$

Explanation:
When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT.
Total number of possible outcomes = 8

Let E be the event of getting exactly two heads.

Then, the favourable outcomes are HHT, THH, and HTH.
Number of favourable outcomes = 3

∴ Probability of getting exactly 2 heads = P(E) =

\frac{Number of favo u rable outcomes}{Total number of possible outcomes} = \frac{3}{8}

Answer:

(b) $\frac{1}{3}$

Explanation:
Number of prizes = 8
Number of blanks = 16
Total number of tickets = 8 +16= 24
∴ P(getting a prize ) = $\frac{8}{24} = \frac{1}{3}$

Answer:

(c) $\frac{4}{5}$

Explanation:
Number of prizes = 6
Number of blanks = 24
Total number of tickets = 6 + 24= 30
∴ P(not getting a prize ) = $\frac{24}{30} = \frac{4}{5}$

Answer:

(c) $\frac{7}{9}$

Explanation:
Total possible outcomes = Total number of marbles

= ( 3 + 2 + 4) = 9

Let E be the event of not getting a white marble.
It means the marble can be either blue or red but not white.
Number of favourable outcomes = (3 + 4) = 7 marbles

∴ P(not getting a white marble ) =

\frac{7}{9}

Answer:

(b) $\frac{3}{5}$

Explanation:
Total possible outcomes = Total number of balls = ( 4 + 6) = 10
Number of black balls = 6
∴ P (getting a black ball ) = $\frac{6}{10} = \frac{3}{5}$

Answer:

(c) $\frac{13}{15}$

Explanation:
Total possible outcomes = Total number of balls

= ( 8 + 2 + 5) = 15

Total number of non-black balls = (8 + 5) = 13
∴ P (getting a ball that is not black) =

\frac{13}{15}

Answer:

(c) $\frac{1}{3}$

Explanation:
Total possible outcomes = Total number of balls

= ( 3 + 4 + 5) = 12

Total number of balls that are non-black and non-white = 4
∴ P (getting a ball that is neither black nor white) =

\frac{4}{12} = \frac{1}{3}

Answer:

(b) $\frac{1}{26}$

Explanation:
Total number of all possible outcomes = 52
Number of black kings = 2
∴ P( getting a black king) = $\frac{2}{52} = \frac{1}{26}$

Answer:

(a) $\frac{1}{13}$

Explanation:
Total number of all possible outcomes= 52
Number of queens = 4
∴ P( getting a queen) = $\frac{4}{52} = \frac{1}{13}$

Answer:

(c) $\frac{3}{13}$

Explanation:
Total number of all possible outcomes= 52
Number of face cards ( 4 kings + 4 queens + 4 jacks) = 12
∴ P( getting a face card) = $\frac{12}{52} = \frac{3}{13}$

Answer:

(b) $\frac{3}{26}$

Explanation:
Total number of all possible outcomes= 52
Number of black face cards ( 2 kings + 2 queens + 2 jacks) = 6
∴ P( getting a black face card) = $\frac{6}{52} = \frac{3}{26}$

Answer:

(c) $\frac{1}{13}$

Explanation:
Total number of all possible outcomes = 52
Number of 6 in a deck of 52 cards = 4
∴ P( getting a 6) = $\frac{4}{52} = \frac{1}{13}$

Rs Aggarwal 2019 for Class 10 Math Chapter 19 - Probability

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