Probability — Exercise 27

A coin is tossed once.

(i) Describe the sample space S.

(ii) Find the probability of getting a tail.

Answer

(i) When a coin is tossed, we get either Head (H) or Tail (T).

∴ Sample space (S) = {H, T}

⇒ n(S) = 2

Hence, sample space = {H, T}.

(ii) Let E be the event of getting a tail.

Then, E = {T}

⇒ n(E) = 1

∴ P(getting a tail) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{2}$

Hence, probability of getting a tail $\dfrac{1}{2}$.

Two coins are tossed simultaneously. Describe the sample space S. Find the probability of getting:

(i) two heads

(ii) at least one head

(iii) at most one head

(iv) exactly one head

(v) no head

Answer

When you toss two coins simultaneously, we get either both heads, first head second tail, first tail second head, both tails.

S = {HH, HT, TH, TT}

Total number of outcomes = 4

(i) Two heads

Number of favorable outcomes (two heads) = 1 (HH)

∴ P(getting two heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}$

Hence, the probability of getting two heads is $\dfrac{1}{4}$.

(ii) At least one head

Number of favorable outcomes (Getting at least one head) = 3 (HT, TH and HH)

∴ P(getting at least one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, the probability of getting at least one head is $\dfrac{3}{4}$.

(iii) At most one head

Number of favorable outcomes (Getting at most one head) = 3 (HT, TH and TT)

∴ P(getting at most one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{4}$

Hence, the probability of getting at most one head is $\dfrac{3}{4}$.

(iv) Exactly one head

Number of favorable outcomes (Getting exactly one head) = 2 (HT, TH)

∴ P(getting exactly one head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{4} = \dfrac{1}{2}.$

Hence, the probability of getting exactly one head is $\dfrac{1}{2}$.

(v) No head

Number of favorable outcomes (Getting no head) = 1 (TT)

∴ P(getting no head) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{4}.$

Hence, the probability of getting no head is $\dfrac{1}{4}$.

Three coins are tossed simultaneously. Describe the sample space S. Find the probability of getting:

(i) at most 2 heads

(ii) at least 2 heads

(iii) exactly 2 heads

Answer

When you toss three coins simultaneously.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total number of outcomes = 8

(i) at most 2 heads

Number of favorable outcomes (Getting at most 2 heads) = 7 (HHT, HTH, THH, HTT, THT, TTH, TTT)

∴ P(getting at most 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{8}.$

Hence, the probability of getting at most 2 heads is $\dfrac{7}{8}$.

(ii) at least 2 heads

Number of favorable outcomes (Getting at least 2 heads) = 4 (HHT, HTH, THH, HHH)

∴ P(getting at least 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{8} = \dfrac{1}{2}.$

Hence, the probability of getting at least 2 heads is $\dfrac{1}{2}$.

(iii) exactly 2 heads

Number of favorable outcomes (Getting exactly 2 heads) = 3 (HHT, HTH, THH)

∴ P(getting exactly 2 heads) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{8} .$

Hence, the probability of getting exactly 2 heads is $\dfrac{3}{8}$.

A die is thrown once. What is the probability of getting:

(i) an odd number

(ii) a number greater than 4

(iii) a number less than 5

(iv) the number 5

Answer

In a single throw of die,

Sample space = {1, 2, 3, 4, 5, 6}.

(i) Let A be the event of getting an odd number, then

A = {1, 3, 5}.

∴ The number of favourable outcomes to the event A = 3.

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{6} = \dfrac{1}{2} .$

Hence, the probability of getting an odd number is $\dfrac{1}{2}.$

(ii) Let B be the event of getting a number greater than 4, then

B = {5, 6}.

∴ The number of favourable outcomes to the event B = 2.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{6} = \dfrac{1}{3} .$

Hence, the probability of getting a number greater than 4 is $\dfrac{1}{3}.$

(iii) Let C be the event of getting a number less than 5, then

C = {1, 2, 3, 4}.

∴ The number of favourable outcomes to the event C = 4.

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{6} = \dfrac{2}{3} .$

Hence, the probability of getting a number less than 5 is $\dfrac{2}{3}.$

(iv) Let D be the event of getting the number 5, then

D = {5}.

∴ The number of favourable outcomes to the event D = 1.

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{6}.$

Hence, the probability of getting the number 5 is $\dfrac{1}{6}.$

Two dice are thrown simultaneously. Find the probability of getting:

(i) 10 as the sum of two numbers that turn up

(ii) a doublet of even numbers

(iii) a total of at least 10

(iv) a multiple of 3 as the sum of two numbers that turn up

Answer

(i) Let A be the event of 10 as the sum of two numbers that turn up, then

A = {(4, 6), (5, 5), (6, 4)}.

∴ The number of favourable outcomes to the event A = 3.

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of 10 as the sum of two numbers that turn up is $\dfrac{1}{12}.$

(ii) Let B be the event of a doublet of even numbers, then

B = {(2, 2), (4, 4), (6, 6)}.

∴ The number of favourable outcomes to the event B = 3.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{36} = \dfrac{1}{12}.$

Hence, the probability of getting a doublet of even numbers is $\dfrac{1}{12}.$

(iii) Let C be the event of a total of at least 10, then

C = {(4, 6), (5, 5) , (6, 4), (5, 6), (6, 5), (6, 6)}.

∴ The number of favourable outcomes to the event C = 6.

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{36} = \dfrac{1}{6}.$

Hence, the probability of getting a total of at least 10 is $\dfrac{1}{6}.$

(iv) Let D be the event of a multiple of 3 as the sum of two numbers that turn up , then

D = {(1, 2), (2, 1), (1, 5), (2, 4), (3, 3) , (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}.

∴ The number of favourable outcomes to the event D = 12.

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{36} = \dfrac{1}{3}.$

Hence, the probability of getting a multiple of 3 as the sum of two numbers that turn up is $\dfrac{1}{3}.$

A box of 160 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is:

(i) defective?

(ii) non-defective?

Answer

Given,

Total number of bulbs = 160

Number of defective bulbs = 12

Number of non defective bulbs = 160 - 12 = 148

(i) Let A be the event of taking out defective bulb , then

∴ The number of favourable outcomes to the event A = 12.

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{160} = \dfrac{3}{40}.$

Hence, the probability of taking out a defective bulb is $\dfrac{3}{40}.$

(ii) Let B be the event of taking out non-defective bulb , then

∴ The number of favourable outcomes to the event B = 148.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{148}{160} = \dfrac{37}{40}.$

Hence, the probability of taking out a non-defective bulb is $\dfrac{37}{40}.$

A box contains 16 cards bearing numbers 1, 2, 3, 4, …, 15, 16 respectively. A card is drawn at random from the box. What is the probability that the number on the card is:

(i) an odd number?

(ii) a prime number?

(iii) a number divisible by 3?

(iv) a number not divisible by 4?

Answer

Given,

Total number of outcomes = 16

(i) Let A be the event of drawing an odd number , then

A = {1, 3, 5, 7, 9, 11, 13, 15}

∴ The number of favourable outcomes to the event A = 8.

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{8}{16} = \dfrac{1}{2}.$

Hence, the probability of drawing an odd number is $\dfrac{1}{2}.$

(ii) Let B be the event of drawing a prime number, then

B = {2, 3, 5, 7, 11, 13}

∴ The number of favourable outcomes to the event B = 6.

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{16} = \dfrac{3}{8}.$

Hence, the probability of drawing a prime number is $\dfrac{3}{8}.$

(iii) Let C be the event of drawing a number divisible by 3, then

C = {3, 6, 9, 12, 15}

∴ The number of favourable outcomes to the event C = 5.

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{16}$

Hence, the probability of drawing a number divisible by 3 is $\dfrac{5}{16}.$

(iv) Let D be the event of drawing a number not divisible by 4, then

D = {1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15}

∴ The number of favourable outcomes to the event D = 12.

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{16} = \dfrac{3}{4}$

Hence, the probability of drawing a number not divisible by 4 is $\dfrac{3}{4}.$

Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is:

(i) a prime number

(ii) a number divisible by 4

(iii) a number that is a multiple of 6

(iv) an odd number

Answer

Given,

Cards in the bag:

Sample space = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}.

Total number of outcomes = 10

(i) Let A be the event of drawing a card with prime number, then

A = {2}

∴ The number of favourable outcomes to the event A = 1

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{10}$

Hence, the probability of drawing a card with prime number is $\dfrac{1}{10}$.

(ii) Let B be the event of drawing a card with number divisible by 4, then

B = {4, 8, 12, 16, 20}

∴ The number of favourable outcomes to the event B = 5

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$

Hence, the probability of drawing a card with number divisible by 4 is $\dfrac{1}{2}$.

(iii) Let C be the event of drawing a card with multiple of 6, then

C = {6, 12, 18}

∴ The number of favourable outcomes to the event C = 3

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{10}$

Hence, the probability of drawing a card with multiple of 6 is $\dfrac{3}{10}$.

(iv) Let D be the event of drawing a card with an odd number, then

D = ∅

∴ The number of favourable outcomes to the event D = 0

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{0}{10} = 0$

Hence, the probability of drawing a card with an odd number is 0.

A box contains 15 balls bearing numbers 1, 2, 3, …, 14, 15 respectively. A ball is drawn at random from the box. Find the probability that the number on the ball is:

(i) an even number

(ii) a number divisible by 5

(iii) the number 6

(iv) a number lying between 8 and 12

(v) a number greater than 9

(vi) a number less than 6

Answer

Given,

Balls in the box are numbered

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Total number of outcomes = 15

(i) Let A be the event of getting an even number, then

A = {2, 4, 6, 8, 10, 12, 14}

∴ The number of favourable outcomes to the event A = 7

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{15}$

Hence, the probability of getting an even number is $\dfrac{7}{15}$.

(ii) Let B be the event of getting a number divisible by 5, then

B = {5, 10, 15}

∴ The number of favourable outcomes to the event B = 3

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15} = \dfrac{1}{5}$

Hence, the probability of getting a number divisible by 5 is $\dfrac{1}{5}$.

(iii) Let C be the event of getting the number 6, then

C = {6}

∴ The number of favourable outcomes to the event C = 1

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{15}$

Hence, the probability of getting the number 6 is $\dfrac{1}{15}$.

(iv) Let D be the event of getting a number between 8 and 12, then

D = {9, 10, 11}

∴ The number of favourable outcomes to the event D = 3

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15} = \dfrac{1}{5}$

Hence, the probability of getting a number lying between 8 and 12 is $\dfrac{1}{5}$.

(v) Let E be the event of getting a number greater than 9, then

E = {10, 11, 12, 13, 14, 15}

∴ The number of favourable outcomes to the event E = 6

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{15} = \dfrac{2}{5}$

Hence, the probability of getting a number greater than 9 is $\dfrac{2}{5}$.

(vi) Let F be the event of getting a number less than 6, then

F = {1, 2, 3, 4, 5}

∴ The number of favourable outcomes to the event F = 5

∴ P(F) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{15} = \dfrac{1}{3}$

Hence, the probability of getting a number less than 6 is $\dfrac{1}{3}$.

There are 25 discs numbered 1 to 25. They are put in a closed box and shaken thoroughly. A disc is drawn at random from the box. Find the probability that the number on the disc is:

(i) an odd number

(ii) divisible by 2 and 3 both

(iii) a number less than 16

Answer

Given,

There are 25 discs numbered from 1 to 25.

Sample space:S={1, 2, 3, …, 25}

Total number of outcomes = 25

(i) Let A be the event of getting an odd number, then

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}

∴ The number of favourable outcomes to the event A = 13

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{13}{25}$

Hence, the probability of getting an odd number is $\dfrac{13}{25}$.

(ii) Let B be the event of getting a number divisible by both 2 and 3, then

B = {6, 12, 18, 24}

∴ The number of favourable outcomes to the event B = 4

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{25}$

Hence, the probability of getting a number divisible by both 2 and 3 is $\dfrac{4}{25}$.

(iii) Let C be the event of getting a number less than 16, then

C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

∴ The number of favourable outcomes to the event C = 15

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{15}{25} = \dfrac{3}{5}$

Hence, the probability of getting a number less than 16 is $\dfrac{3}{5}$.

In a class of 40 students, there are 16 boys and the rest are girls. From these students, one is selected at random. What is the probability that the selected student is a girl?

Answer

Given,

Total number of students = 40

Number of boys = 16

Let G be the event of selecting a girl,

∴ The number of favourable outcomes to the event G = 24

∴ P(G) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{24}{40} = \dfrac{3}{5}$

Hence, the probability of selecting a girl is $\dfrac{3}{5}$.

A bag contains 8 red, 4 white and 3 black balls. One ball is drawn at random. What is the probability that the ball drawn is:

(i) white?

(ii) red or white?

(iii) neither red nor white?

(iv) not red?

Answer

Given,

Total number of outcomes = 15

(i) Let A be the event of getting white ball, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{15}$

Hence, the probability of getting white ball is $\dfrac{4}{15}$.

(ii) Let B be the event of getting white or red ball, then

∴ The number of favourable outcomes to the event B = 12(white + red balls)

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{15}= \dfrac{4}{5}$

Hence, the probability of getting white or red ball is $\dfrac{4}{5}$.

(iii) Let C be the event of getting neither red nor white ball, then

∴ The number of favourable outcomes to the event C = 3 (black balls)

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{3}{15}= \dfrac{1}{5}$

Hence, the probability of getting neither red nor white ball is $\dfrac{1}{5}$.

(iv) Let D be the event of not getting red ball, then

∴ The number of favourable outcomes to the event D = 7(white + black balls)

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{15}$

Hence, the probability of not getting red ball is $\dfrac{7}{15}$.

A bag contains 6 black, 5 white and 9 green balls. One ball is drawn at random. What is the probability that the ball drawn is:

(i) black?

(ii) not green?

(iii) either white or green?

(iv) neither white nor black?

Answer

Given,

Total number of outcomes = 6 (Black) + 5 (White) + 9 (Green) = 20

(i) Let A be the event of getting black ball, then

∴ The number of favourable outcomes to the event A = 6

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{20} = \dfrac{3}{10}$

Hence, the probability of getting black ball is $\dfrac{3}{10}$.

(ii) Let B be the event of not getting green ball, then

∴ The number of favourable outcomes to the event B = 6 (Black) + 5 (White) = 11

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{11}{20}$

Hence, the probability of not getting green ball is $\dfrac{11}{20}$.

(iii) Let C be the event of getting either white or green ball, then

∴ The number of favourable outcomes to the event C = 5 (White) + 9 (Green) = 14

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{14}{20}= \dfrac{7}{10}$

Hence, the probability of getting either white or green ball is $\dfrac{7}{10}$.

(iv) Let D be the event of getting neither white nor black ball , then

∴ The number of favourable outcomes to the event D = 9(green balls)

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{20}$

Hence, the probability of getting neither white nor black ball is $\dfrac{9}{20}$.

One card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of drawing:

(i) an ace

(ii) a 5 of a red suit

(iii) a black queen

(iv) a jack of spades

(v) a 10 of hearts

(vi) a face card

Answer

Given,

Total number of outcomes = 52

(i) Let A be the event of getting an ace, then

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{52} = \dfrac{1}{13}$

Hence, the probability of getting an ace is $\dfrac{1}{13}$.

(ii) Let B be the event of getting 5 of red suit, then

∴ The number of favourable outcomes to the event B = 2

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$

Hence, the probability of getting 5 of red suit is $\dfrac{1}{26}$.

(iii) Let C be the event of getting a black queen, then

∴ The number of favourable outcomes to the event C = 2 (one of club and one of spade)

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{52} = \dfrac{1}{26}$

Hence, the probability of getting a black queen is $\dfrac{1}{26}$.

(iv) Let D be the event of getting a jack of spades, then

∴ The number of favourable outcomes to the event D = 1

∴ P(D) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{52}$

Hence, the probability of getting a jack of spades is $\dfrac{1}{52}$.

(v) Let E be the event of getting a 10 of hearts, then

∴ The number of favourable outcomes to the event E = 1

∴ P(E) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{1}{52}$

Hence, the probability of getting a 10 of hearts is $\dfrac{1}{52}$.

(vi) Let F be the event of getting a face card, then

There are 12 face cards in a deck.

∴ The number of favourable outcomes to the event F = 12

∴ P(F) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

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

A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability that the card drawn is:

(i) either a king or a queen

(ii) neither a king nor a queen

Answer

Given,

Total number of outcomes = 52

(i) Let A be the event of getting either a king or a queen, then

∴ The number of favourable outcomes to the event A = 4 (Kings) + 4 (Queens) = 8

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{8}{52} = \dfrac{2}{13}$

Hence, the probability of getting either a king or a queen is $\dfrac{2}{13}$.

(ii) Let B be the event of getting neither a king nor a queen, then

∴ The number of favourable outcomes to the event B = Total cards - (Kings + Queens) = 44

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{44}{52} = \dfrac{11}{13}$

Hence, the probability of getting neither a king nor a queen is $\dfrac{11}{13}$.

If the probability of winning a game is 0.6, what is the probability of losing the game?

Answer

We know that,

P(E) + P(not E) = 1

Given,

Probability of winning game = 0.6

P(winning) + P(losing) = 1

P(losing) = 1 - 0.6 = 0.4

Hence, the probability of losing game is 0.4

Fill in the blanks:

(i) The probability of a sure event is __________.

(ii) The probability of an impossible event is __________.

(iii) For an event E, we have P(E) + P(not E) = __________.

(iv) For an event E, we have () ≤ P(E) ≤ ().

Answer

Fill in the blanks:

(i) The probability of a sure event is 1.

(ii) The probability of an impossible event is 0.

(iii) For an event E, we have P(E) + P(not E) = 1.

(iv) For an event E, we have 0 ≤ P(E) ≤ 1.

Sixteen cards are labelled as a, b, c, …, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:

(i) a vowel

(ii) a consonant

(iii) none of the letters of the word “median”

Answer

Given,

Sample space S = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p}

Total number of outcomes = 16

(i) Let A be the event of getting a vowel, then

A = {a, e, i, o}

∴ The number of favourable outcomes to the event A = 4

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{4}{16} = \dfrac{1}{4}$

Hence, the probability of getting a vowel is $\dfrac{1}{4}$.

(ii) Let B be the event of getting a consonant, then

B = {b, c, d, f, g, h, j, k, l, m, n, p}

∴ The number of favourable outcomes to the event B = 12

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{12}{16} = \dfrac{3}{4}$

Hence, the probability of getting a consonant is $\dfrac{3}{4}$.

(iii) Let C be the event of not getting letters of word median, then

C = {b, c, f, g, h, j, k, l, o, p}

∴ The number of favourable outcomes to the event C = 10

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{16} = \dfrac{5}{8}$

Hence, the probability of not getting letters of word median is $\dfrac{5}{8}$.

The following letters A, D, M, N, O, S, U, Y of the English alphabet are written on separate cards and put in a box. The cards are well shuffled and one card is drawn at random. What is the probability that the card drawn is a letter of the word,

(i) MONDAY?

(ii) which does not appear in MONDAY?

(iii) which appears both in SUNDAY and MONDAY?

Answer

Letters written on cards = {'A', 'D', 'M', 'N', 'O', 'S', 'U', 'Y'}

No. of cards = 8

(i) Letters of the word MONDAY present in the cards = {'M', 'O', 'N', 'D', 'A', 'Y'}

Probability that the card drawn is a letter of the word MONDAY

= $\dfrac{\text{Letters of MONDAY present}}{\text{No. of cards}} = \dfrac{6}{8} = \dfrac{3}{4}$.

Hence, required probability = $\dfrac{3}{4}$.

(ii) Letters of the word not present in MONDAY = {'S', 'U'}

Probability that the card drawn is not a letter of the word MONDAY

= $\dfrac{\text{Letters not present in MONDAY}}{\text{No. of cards}} = \dfrac{2}{8} = \dfrac{1}{4}$.

Hence, required probability = $\dfrac{1}{4}$.

(iii) Letters of the word present in SUNDAY and MONDAY are {'N', 'D', 'A', 'Y'}.

Probability that the card drawn has a letter which appears both in SUNDAY and MONDAY

= $\dfrac{\text{Letters present in both MONDAY and SUNDAY}}{\text{No. of cards}} = \dfrac{4}{8} = \dfrac{1}{2}$.

Hence, required probability = $\dfrac{1}{2}$.

A bag contains 25 cards, numbered through 1 to 25. A card is drawn at random. What is the probability that the number on the card drawn is:

(i) a multiple of 5

(ii) a perfect square

(iii) a prime number

Answer

Given,

Sample space = {1, 2, 3, 4, 5, ....., 25}

Total number of outcomes = 25

(i) Let A be the event of getting a multiple of 5, then

A = {5, 10, 15, 20, 25}

∴ The number of favourable outcomes to the event A = 5

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

Hence, the probability of getting a multiple of 5 is $\dfrac{1}{5}$.

(ii) Let B be the event of getting a perfect square, then

B = {1, 4, 9, 16, 25}

∴ The number of favourable outcomes to the event B = 5

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{25} = \dfrac{1}{5}$

Hence, the probability of getting a perfect square is $\dfrac{1}{5}$.

(iii) Let C be the event of getting a prime number, then

C = {2, 3, 5, 7, 11, 13, 17, 19, 23}

∴ The number of favourable outcomes to the event C = 9

∴ P(C) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{9}{25}$

Hence, the probability of getting a prime number is $\dfrac{9}{25}$.

A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random. What is the probability that the ball drawn is a red ball?

Answer

Total number of outcomes = 5 white + 2 red + 3 black balls = 10

Let A be the event of getting a red ball.

The number of favorable outcomes to the event A = 2

∴ P(A) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}$

Hence, the probability of drawing a red ball is $\dfrac{1}{5}$.

A letter of the word SECONDARY is selected at random. What is the probability that the letter selected is not a vowel?

Answer

Given, Word = SECONDARY The letters are {S, E, C, O, N, D, A, R, Y}

Total number of outcomes = 9

The letters that are consonants = {S, C, N, D, R, Y}.

Let B be the event of selecting a letter that is not a vowel.

The number of favorable outcomes to the event B = 6

∴ P(B) = $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{6}{9} = \dfrac{2}{3}$

Hence, the probability that the letter selected is not a vowel is $\dfrac{2}{3}$.

Ms. Sushmita went to a fair and participated in a game. The game consisted of a box having number cards with numbers from 01 to 30. The three prizes were as per the given table:

Find the probability of winning a:

(i) Wall Clock

(ii) Water Bottle

(iii) Purse

Answer

Given,

Total number of outcomes = 30

(i) Given,

Numbers that are perfect squares (between 1 to 30) are 1, 4, 9, 16, 25 (5 numbers).

∴ No. of favourable outcomes = 5 perfect squares

P(perfect square) $=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{30} = \dfrac{1}{6}$.

Hence, probability of numbers that are perfect squares = $\dfrac{1}{6}$.

(ii) Given,

Even numbers that are also multiples of 3 (between 1 to 30) are 6, 12, 18, 24, 30 (5 numbers)

∴ No. of favourable outcomes = 5 Even numbers that are also multiples of 3

P(even and multiple of 3) $=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{5}{30} = \dfrac{1}{6}$

Hence, probability of numbers that are multiples of 3 and even number = $\dfrac{1}{6}$.

(iii) Given,

Prime numbers (1 to 30) are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (10 numbers)

∴ No. of favourable outcomes = 10 Prime numbers

P(prime)

$=\dfrac{\text{No of favourable outcomes}}{\text{Total number of outcomes}} = \dfrac{10}{30} = \dfrac{1}{3}$

Hence, probability of prime numbers = $\dfrac{1}{3}$.

A box containing cards numbered between 0 and 100 are shuffled and a card is picked at random. Find the probability of getting a card which is:

(i) divisible by 6.

(ii) not divisible by 6.

Answer

The cards are numbered from 0 to 100.

Total number of cards = 101 (as there is no. zero card also)

(i) Favorable outcome = Getting a card having a number which is divisible by 6.

{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

Probability $=\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{17}{101}$.

Hence, probability of getting a card which is divisible by 6 = $\dfrac{17}{101}$.

(ii) Favorable outcome = Getting a card having a number which is not divisible by 6.

P(getting a card not divisible by 6) = 1 - P(getting a card divisible by 6)

= $1 - \dfrac{17}{101}$

= $\dfrac{101 - 17}{101}$

= $\dfrac{84}{101}$.

Hence, probability of getting a card which is not divisible by 6 = $\dfrac{84}{101}$.

There are some red, green and white marbles in a box. One marble is picked up at random from this box. If the probability of picking up a red marble is $\dfrac{2}{9}$ and that of picking up a green marble is $\dfrac{4}{9}$ then find the :

(i) probability of picking up a white marble.

(ii) number of green marbles, if total number of marbles is 54.

(iii) probability of not picking up a red marble.

Answer

(i) The sum of the probabilities of all possible outcomes is equal to 1.

The possible outcomes are picking a red, green, or white marble.

P(red) + P(green) + P(white) = 1

Given,

P(red) = $\dfrac{2}{9}$

P(green) = $\dfrac{4}{9}$

$\Rightarrow \dfrac{2}{9} + \dfrac{4}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow \dfrac{6}{9} + P(\text{white}) = 1 \\[1em] \Rightarrow P(\text{white}) = 1 - \dfrac{6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{9 - 6}{9} \\[1em] \Rightarrow P(\text{white}) = \dfrac{3}{9} = \dfrac{1}{3}.$

Hence, probability of picking up a white marble = $\dfrac{1}{3}$.

(ii) Given,

Total number of marbles = 54

P(green) $=\dfrac{\text{No of favorable outcomes}}{\text{Total number of outcomes}}$.

$\Rightarrow \dfrac{4}{9} = \dfrac{\text{Number of green marbles}}{54} \\[1em] \Rightarrow \dfrac{4}{9} \times 54 = \text{Number of green marbles} \\[1em] \Rightarrow 4 \times 6 = \text{Number of green marbles} \\[1em] \Rightarrow \text{Number of green marbles} = 24.$

Hence, number of green marbles is 24.

(iii) Given,

P(red) = $\dfrac{2}{9}$

P(not picking a red) = 1 - P(red)

= 1 - $\dfrac{2}{9}$

= $\dfrac{9 - 2}{9}$

= $\dfrac{7}{9}$.

Hence, probability of not picking up a red marble is $\dfrac{7}{9}$.