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# Class 11 RD Sharma Solutions – Chapter 33 Probability – Exercise 33.3 | Set 3

### Question 31. A bag contains 5 red, 6 white and 7 black balls. Two balls are drawn at random. Find the probability that both balls are red or black.

Solution:

Clearly the bag has 5 + 6 + 7 = 18 balls and since 2 balls have been drawn at random, number of outcomes in sample space = n{S} = 18C2 = 153

Let E be the event that the two balls chosen are either red or black.

Hence number of outcomes in event E = n{E} = 5C2 + 7C2 = 31

Probability of E = n{E}/n{S} = 31/153

Hence the probability of both balls being red or black is 31/153.

### Question 32. If a letter is chosen at random from the English alphabet, find the probability that it is a:

(i) vowel

Solution:

We know there are 26 letters in English alphabet. Since one letter has been chosen from the 26 letters, number of outcomes in sample space = n{S} = 26C1 = 26

Let V be the event that the chosen letter is a vowel. Obviously there are 5 vowels in the English alphabet.

Hence the number of outcomes favorable to event V = n{V} = 5C1 = 5.

Probability of event V = n{V}/n{S} = 5/26.

Hence the probability of getting a vowel is 5/26.

(ii) consonant

Solution:

Let C be the event that the chosen letter is a consonant. Obviously there are 21 consonants in the English alphabet.

Hence the number of outcomes favorable to event C = n{C} = 21C1 = 21.

Since n{S} = 26, Probability of event C = n{C}/n{S} = 21/26.

Hence the probability of getting a consonant is 21/26.

### Question 33. In a lottery, a person chooses 6 different numbers from 1 to 20, and if these six numbers match with the six numbers fixed by the lottery committee, he wins the prize. What is the probability of winning the game?

Solution:

Since there are 20 numbers, and 6 numbers have to be chosen, the number of outcomes in sample space = n{S} = 20C6 = 38760

Let L be the event that the 6 numbers chosen match with the already decided numbers. We know that there is only 1 set pattern to the numbers that have already been fixed by the lottery committee.

Hence number of outcomes favorable to event L = n{L} = 1.

Probability of event L = n{L}/n{S} = 1/38760.

Hence the probability of winning the lottery is 1/38760.

### Question 34. 20 cards are numbered from 1 to 20. One card is drawn at random. Find the probability that the number on the cards is:

(i) a multiple of 4

Solution:

Since 1 card has been drawn from 20 given cards, number of outcomes in sample space = n{S}

=> 20C1 = 20.

Let Y be the event that the number on the card is a multiple of 4. We know there are 5 multiples of 4 from 1 to 20, (i.e., 4, 8, 12, 16, 20.)

Hence number of outcomes favorable to Y = n{Y} = 5

Probability of event Y = P{Y} = n{Y}/n{S} = 5/20 = 1/4.

Hence the probability of the number on the card being a multiple of 4 is 1/4.

(ii) not a multiple of 4

Solution:

Let Y be the event that the number on the card is a multiple of 4 and Y’ denote the event that it is not a multiple of 4.

We know there are 5 multiples of 4 from 1 to 20, (i.e., 4, 8, 12, 16, 20.)

Hence number of outcomes favorable to Y = n{Y} = 5

Since n{S} = 20, Probability of event Y = n{Y}/n{S} = 5/20 = 1/4.

Thus, P{Y’} = 1 − P{Y} = 1 − 1/4 = 3/4.

Hence the probability of the number on the card not being a multiple of 4 is 3/4.

(iii) odd

Solution:

Let O be the event that the number on the card is odd. We know there are 10 odd numbers from 1 to 20, (i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.)

Hence number of outcomes favorable to O = n{O} = 10

Since n{S} = 20, Probability of event O = P{O} = n{O}/n{S} = 10/20 = 1/2.

Hence the probability of the number on the card being an odd number is 1/2.

(iv) greater than 12

Solution:

Let G be the event that the number on the card is odd. We know there are 8 numbers greater than 12 from 1 to 20, (i.e., 13, 14, 15, 16, 17, 18, 19, 20.)

Hence number of outcomes favorable to G = n{G} = 8

Since n{S} = 20, Probability of event G = P{G} = n{G}/n{S} = 8/20 = 2/5.

Hence the probability of the number on the card being greater than 12 is 2/5.

(v) divisible by 5

Solution:

Let F be the event that the number on the card is divisible by 5. We know there are 4 numbers divisible by 5 from 1 to 20, (i.e., 5, 10, 15, 20.)

Hence number of outcomes favorable to F = n{F} = 4

Since n{S} = 20, Probability of event F = P{F} = n{F}/n{S} = 4/20 = 1/5.

Hence the probability of the number on the card being divisible by 5 is 1/5.

(vi) not a multiple of 6

Solution:

Let Z be the event that the number on the card is a multiple of 6 and Z’ denote the event that it is not a multiple of 6.

We know there are 3 multiples of 6 from 1 to 20, (i.e., 6, 12, 18.)

Hence number of outcomes favorable to Z = n{Z} = 3

Since n{S} = 20, Probability of event Z = n{Z}/n{S} = 3/20

Thus, P{Z’} = 1 − P{Z} = 1 − 3/20 = 17/20.

Hence the probability of the number on the card not being a multiple of 6 is 17/20.

### Question 35. Two dice are thrown. Find the odds:

(i) in favour of getting the sum 4

Solution:

Since a pair of dice has been thrown, number of items in sample space = n{S}= 62 = 36.

Let E be the event of getting a sum of 4 on the dice and E’ be the event of not getting a sum of 4.

E = {(1,3), (2,2), (3,1)}

Hence number of outcomes favorable to event E = n{E} = 3

Probability of event E = P{E} = n{E}/n{S} = 3/36 = 1/12

Now, P{E’} = 1 − P{E} = 1 − 1/12 = 11/12.

Thus odds in favour of getting a sum of 4 = P{E}/P{E’} = 1:11.

(ii) in favour of getting the sum 5

Solution:

Let C be the event of getting a sum of 5 on the dice and E’ be the event of not getting a sum of 5.

C = {(1,4), (2,3), (3,2), (4,1)}

Hence number of outcomes favorable to event C = n{C} = 4

Since n{S} = 36, Probability of event C = P{C} = n{C}/n{S} = 4/36 = 1/9

Now, P{C’} = 1 − P{C} = 1 − 1/9 = 8/9.

Thus odds in favour of getting a sum of 5 = P{C}/P{C’} = 1:8.

(iii) against getting a sum of 6?

Solution:

Let A be the event of getting a sum of 6 on the dice and A’ be the event of not getting a sum of 6.

A = {(1,5), (2,4), (3,3), (4,2), (5,1)}

Hence number of outcomes favorable to event A = n{A} = 5

Since n{S} = 36, Probability of event A = P{A} = n{A}/n{S} = 5/36

Now, P{A’} = 1 − P{A} = 1 − 5/36 = 31/36.

Thus odds against getting a sum of 6 = P{A’}/P{A} = 31:5.

### Question 36. What are the odds in favour of getting:

(i) a spade if the card is drawn from a deck of well- shuffled cards?

Solution:

Let W be the event of getting a spade from a well shuffled deck of cards and W’ be the event of not getting a spade.

We know there are 13 spade cards in a pack of cards.

Hence number of outcomes favorable to W = n{W} = 13

Probability of getting a spade = P{W} = n{W}/n{S} = 13/52 =1/4

Thus, probability of not getting a spade = P{W’} = 1 − P{W} = 1 −1/4 = 3/4

Odds in favour of getting a spade = P{W}/P{W’} = (1/4)/(3/4) = 1:4

(ii) a king?

Solution:

Let K be the event of getting a king from a well shuffled deck of cards and K’ be the event of not getting a king.

We know there are 4 kings in a pack of cards.

Hence number of outcomes favorable to K = n{K} = 13

Probability of getting a king = P{K} = n{K}/n{S} = 4/52 =1/13

Thus, probability of not getting a king = P{K’} = 1 − P{K} = 1 −1/13 = 12/13

Odds in favour of getting a king = P{K}/P{K’} = (1/13)/(12/13) = 1:12

### Question 37. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. Find the probability that:

(i) all are blue

Solution:

Clearly the box has 10 + 20 + 30 = 60 marbles and since 5 marbles have been drawn at random, number of outcomes in sample space = n{S} = 60C5

Let B be the event that all 5 marbles are blue.

Hence number of outcomes favorable to event B = n{B} = 20C5

Probability of B = n{B}/n{S} = 20C5 / 60C5 = 34/11977

Hence the probability of all 5 marbles being blue is 34/11977.

(ii) at least one is green

Solution:

Let G be the event that at least one marble is drawn.

Probability of G = 1 − Probability of drawing no green marble = 1 − 30C5/60C5 = 1 − 117/4484 = 4367/4484

Hence the probability of drawing at least one green marble is 4367/4484.

### Question 38: A box has 6 red marbles numbered 1 through 6 and 4 white marbles numbered 12 through 15. Find the probability that a marble drawn is:

(i) white

Solution:

Since 1 marble has been drawn from a box of 10 marbles, number of outcomes in sample space = n{S} = 10C1 = 10.

Let W be the event that the given marble is white.

Number of outcomes favorable to event W = n{W} = 4C1 = 4.

Probability of W = n{W}/n{S} = 4/10 = 2/5

Hence the probability of the marble being white is 2/5.

(ii) white and odd-numbered

Solution:

Let Q be the event that the marble drawn is white and odd numbered. We know there are 2 odd numbers between 12 and 15, (i.e., 13 and 15).

Number of outcomes favorable to event Q = 2

Since n{S} = 10, Probability of Q = n{Q}/n{S} = 2/10 = 1/5

Hence the probability of the marble being white and odd numbered is 1/5.

(iii) even-numbered

Solution:

Let V be the event that the marble drawn is even numbered.

V = {2, 4, 6, 12, 14}

Number of outcomes favorable to event V = 5

Since n{S} = 10, Probability of V = n{V}/n{S} = 5/10 = 1/2

Hence the probability of the marble being even numbered is 1/2.

(iv) red or even-numbered

Solution:

Let G1 be the event of getting a red marble. We know there are 6 red marbles, hence probability of G1 = P{G1} = 6/10

Let G2 be the event of getting an even numbered marble. We know there are 5 even numbered marbles, hence probability of G2 = P{G2} = 5/10

Now, {G1 ∩ G2} = even numbered red marble = {2, 4, 6}

Thus, n{G1 ∩ G2} = 3 and Probability of {G1 ∩ G2} = P{G1 ∩ G2} = 3/10.

P{G1 ∪ G2} = P{G1} + P{G2} − P{G1 ∩ G2} = 6/10 + 5/10 − 3/10 = 8/10 = 4/5

Hence the probability of getting a red or even numbered marble is 4/5.

### Question 39. A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has:

(i) all boys

Solution:

Since 3 students have been selected at random from a group of 18 students, number of outcomes in sample space = n{S} = 18C3 = 816

Let B be the event that the selected group has all boys.

Hence number of outcomes favorable to B = n{B} =10C3 = 120

Probability of event B = n{B}/n{S} = 120/816 = 5/34

Hence the probability that the group has all boys is 5/34.

(ii) all girls

Solution:

Let G be the event that the selected group has all girls.

Hence number of outcomes favorable to G = n{G} = 8C3 = 56

Since n{S} = 816, Probability of event G = n{G}/n{S} = 56/816 = 7/102

Hence the probability that the group has all girls is 7/102.

(iii) 1 boy and 2 girls

Solution:

Let H be the event that the selected group has 1 boy and 2 girls.

Hence number of outcomes favorable to H = n{H} = 10C1 × 8C2 = 280

Since n{S} = 816, Probability of event H = n{H}/n{S} = 280/816 = 35/102

Hence the probability that the group has 1boy and 2 girls is 35/102.

(iv) at least 1 girl

Solution:

Let C be the event that the selected group has at least 1 girl.

Probability of having at least 1 girl

=> 1 − P{no girl in the group} = 1 − P{all boys} = 1 − 10C3/18C3 = 1 − 5/34 = 29/34

Hence probability of having at least 1 girl is 29/34.

(v) at most 1 girl

Solution:

Let T be the event that the selected group has at most 1 girl.

T = {0 girl, 1 girl}

Thus, n{T} = 8C0 × 10C3 + 8C1 × 10C2 = 480

Since n{S} = 816, Probability of event T = n{T}/n{S} = 480/816 =10/17

Hence probability of having at most 1 girl is 10/17.

### Question 40. Five cards are drawn from a well-shuffled pack of 52 cards. Find the probability that all the 5 cards are hearts.

Solution:

Since 5 cards have been drawn from a well shuffled pack of 52 cards, number of outcomes in sample space = n{S} = 52C5 = 2598960.

Let H be the event that all 5 cards drawn are hearts. We know there are 13 hearts in a pack of 52 cards.

Hence number of outcome favorable to event H = n{H} = 13C5 = 1287.

Probability of event H = P{H} = n{H}/n{S} = 1287/2598960 = 33/66640

Hence the probability of getting 5 hearts is 33/66640.

### Question 41. A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that:

(i) both the tickets have prime numbers on them

Solution:

Since 2 tickets have been drawn from a bag containing 20 tickets, number of outcomes in sample space = n{S} = 20C2 = 190.

Let N be the event that both the tickets have prime numbers on them. We know there are 8 prime numbers between 1 to 20, (i.e., 2, 3, 5, 7, 9, 11, 13, 17, 19).

Hence number of outcomes favorable to event N = 8C2 = 28

Probability of event N = P{N} = n{N}/n{S} = 28/190 = 14/95

Hence the probability of both the tickets having prime numbers on them is 14/95.

(ii) on one there is a prime number and on the other there is a multiple of 4

Solution:

Let T be the event that on one ticket there is a prime number and on the other there is a multiple of 4.

We know there are 8 prime numbers 5 multiples of 4 from 1 to 20.

Hence number of outcomes in event T = n{T} = 8C1 × 5C1 = 8 × 5 = 40.

Since n{S} = 190, Probability of event T = P{T} = 40/190 = 4/19

Thus the probability of getting a prime number on one ticket and a multiple of 4 on the other is 4/19.

### Question 42. An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random. Find the probability that:

(i) both the balls are red

Solution:

Clearly the urn has 7 + 5 + 3 = 15 balls and since 2 balls have been drawn at random, number of outcomes in sample space = n{S} = 15C2 =105.

Let R be the event that the two balls chosen are red.

Hence number of outcomes in event R = n{R} = 3C2 = 3

Probability of R = n{R}/n{S} = 3/105 = 1/35.

Hence the probability of both balls being red is 1/35.

(ii) one ball is red and the other is black

Solution:

Let X be the event that one ball is red and the other is black.

Hence number of outcomes in event X = n{X} = 3C1 × 5C1 = 15

Since n{S} = 105, Probability of X = n{X}/n{S} = 15/105 = 1/7.

Hence the probability of one ball being red and the other being black is 1/7.

(iii) one ball is white

Solution:

Let D be the event that one ball is white. It means that the other ball can be from the 8 remaining balls, excluding the white balls.

Hence number of outcomes in event D = n{D} = 7C1 × 8C1 = 56

Since n{S} = 105, Probability of D = n{D}/n{S} = 56/105 = 8/15.

Hence the probability of one ball being white is 8/15.

### Question 43. A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

Solution:

Since a pair of dice has been thrown, number of items in sample space = n{S}= 62 = 36.

Let E be the event that B throws a higher number than 9, which is either 10 or 11 or 12.

Hence, E = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

Hence number of outcomes favorable to E = 6

Probability of event E = n{E}/n{S} = 6/36 = 1/6

Hence the probability of B throwing a higher number is 1/6.

### Question 44. In a hand at Whist, what is the probability that 4 kings are held by a specified player?

Solution:

We know that in a hand at Whist, each given player has a pack of 13 cards.

Hence number of outcomes in sample space = n{S} = 52C13

Let K denote the event that a player has 4 kings. It implies that the player must have 9 cards from the pack of the rest of 48 cards(excluding the 4 kings already with the player.)

Number of outcomes favorable to event K = n{K} = 4C4 × 48C9 = 4 × 48C9

Probability of K = P{K} = n{K}/n{S} = 4 × 48C9 / 52C13 = 11/4165

Hence the probability that a specific player has all the 4 hearts is 11/4165.

### Question 45. Find the probability that in a random arrangement of letters of the word ‘UNIVERSITY’, the two I’s do not come together.

Solution:

We know there are 10 letters in the word ‘UNIVERSITY’. Hence number of outcomes in sample space = n{S} = 10!

Let W denote the event that both I’s do not come together and let W’ denote the event that both I’s come together.

Hence number of outcomes favorable to event W’ = n{W’} = 2 × 9!

Thus, Probability of W’ = P{W’} = n{W’}/n{S} = 2 × 9!/10! = 2/10 = 1/5

Now, P{W} = 1 − P{W’} = 1 − 1/5 = 4/5

Hence the probability of not getting the two I’s together is 4/5.

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