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# Class 11 RD Sharma Solutions – Chapter 33 Probability – Exercise 33.3 | Set 2

• Last Updated : 08 May, 2021

### Question 16. A bag contains 7 white, 5 black, and 4 red balls. If two balls are drawn at random, find the probability that:

(i) both the balls are white

Solution:

There are 7+5+4 =16 balls in the bag and since 2 balls have been drawn at random, the total number of outcomes in sample space= n{S}= 16C2 = 120.

Let W be the event of getting both white balls.

Since there are 7 white balls, number of outcomes in W= n{W} = 7C2 = 21.

Probability of event= P{W} = n{W}/n{S} = 21/120 = 7/40

Hence probability of both balls being white is 7/40.

(ii) one ball is black and the other red

Solution:

Let A be the event of getting one black and one red ball.

Hence number of outcomes in A= n{A}= 5C1 × 4C1 = 20.

Since n{S}= 120, Probability of A= P{A}= n{A}/n{S}= 20/120= 1/6

Hence probability of one ball being black and the other being red is 1/6.

(iii) both the balls are of the same color

Solution:

Let O be the event of getting both balls of same color.

Hence number of outcomes in O= n{O}= 7C2 +5C2 + 4C2 = 37.

Since n{S}= 120, Probability of O= P{O}= n{O}/n{S}= 37 / 120.

Hence probability of both the balls being of the same color is 37/120.

### Question 17. A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that:

(i) one is red and two are white

Solution:

There are 6+4+8 = 18 balls in the bag and since 3 balls have been drawn at random, the total number of outcomes in sample space= n{S}= 18C3 = 816.

Let Y be the event of getting one red and two white balls.

Hence number of outcomes in Y = n{Y} = 6C1 × 4C2 = 36.

Probability of event= P{Y} = n{Y}/n{S} = 36/816 = 3/68

Hence probability of one ball being red and the other two being white is 3/68.

(ii) two are blue and one is red

Solution:

Let L be the event of getting two blue and one red balls.

Hence number of outcomes in L= n{L}= 8C2 × 6C1=168.

Since n{S}= 120, Probability of L= P{L}= n{L}/n{S}= 168/816= 7/34

Hence probability of one ball being red and the other two being blue is 7/34.

(iii) one is red

Solution:

Let X be the event that one of the balls must be red.

Hence number of outcomes in X= n{X}= 6C1 × 4C1 × 8C1 + 6C1 × 4C2 + 6C1 × 8C2 = 396.

Since n{S}= 120, Probability of X= P{X}= n{X}/n{S} = 396/816 = 33/68

Hence probability of one ball being red is 33/68.

### Question 18. Five cards are drawn from a pack of 52 cards. What is the chance that these 5 will contain:

(i) just one ace

Solution:

Since five cards are drawn at random from a pack of 52 cards, total number of outcomes in sample space= n{S}= 52C5 = 2598960

Let E be the event that exactly only one ace is present.

Hence number of outcomes in E= n{E}= 4C1 × 48C4 = 778320

Probability of event= P{E} = n{E}/n{S}= 778320/2598960= 3243/10829

Hence probability of getting just one ace is 3243/10829.

(ii) at least one ace

Solution:

Let K be the event that at least one ace is present in the cards drawn from the pack of 52 cards.

Hence event K= {1 or 2 or 3 or 4 ace(s)}

Hence n{K} = 4C1 × 48C4 + 4C2 × 48C3 + 4C3 × 48C2 + 4C4 × 48C1 = 886656

Probability of event K= P{K} = n{K}/n{S}= 886656/2598960= 18472/54145

Hence probability of getting at least one ace is 18472/54145.

### Question 19. The face cards are removed from a full pack. Out of the remaining 40 cards, 4 are drawn at random. What is the probability that they belong to different suits?

Solution:

Since there are 12 face cards(4 queens, 4 kings, 4 jacks and 4 aces) in a pack of 52 cards, the total number of cards left are 40.

Out of these 40 cards, the number of ways of choosing 4 cards= n{S}= 40C4 = 91390

Let E be the event that 4 cards belong to different suit.

Hence number of outcomes in E= n{E}= 10C1 × 10C1 × 10C1 × 10C1 = 10000

Probability of event E= P{E} = n{E}/n{S}= 10000/91390= 1000/9139

Hence the probability of the cards belonging to different suits is 1000/9139.

### Question 20. There are four men and six women on the city councils. If one council member is selected for a committee at random, how likely is that it is a woman?

Solution:

There are four men and six women on the city councils. Hence total number of people on the council is 10.

Since 1 council member is selected at random number of outcomes in sample space= n{S}= 10C1 = 10.

Let E be the event that it is a woman.

Hence number of outcomes in E= n{E} = 6C1 = 6

Probability of event E= P{E} = n{E}/n{S} = 6/10 = 3/5

Hence the probability of the member chosen at random being a woman is 3/5.

### Question 21. A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:

(i) all 10 are defective

Solution:

Since ten bulbs have been drawn at random for inspection from a bag of 100 bulbs, total possible outcomes in sample space= n{S}= 100C10

Let E be the event that all ten bulbs are defective. We know 20 bulbs are defective.

Hence number of outcomes in E= n{E}= 20C10

Probability of event E= P{E} = n{E}/n{S}= 20C10 / 100C10

Hence the probability of all 10 bulbs being defective is 20C10 / 100C10.

(ii) all 10 are good

Solution:

Let U be the event that all ten good bulbs are selected.

We know that 20 bulbs are defective. So, number of good bulbs= 100− 20= 80.

Since all 10 bulbs are good, number of outcomes in event U= n{U} = 80C10

Since n{S}= 100C10, Probability of event U=P{U}= n{U}/n{S}= 80C10 / 100C10

Hence the probability of all balls being good is 80C10 / 100C10.

(iii) at least one is defective

Solution:

Let R be the event that at least one bulb is defective. Since there are 10 defective bulbs, let us assign a number to each number from 1 to 10.

Thus, R= {1,2,3,4,5,6,7,8,9,10}

Let R′ be the event that none of the bulb is defective

n{R’}= 80C10

Since n{S}= 100C10 , Probability of R’= P{R’}= n{R’}/n{S}= 80C10 / 100C10

So, P{R}= 1 – P{R’}= 1 – 80C10 / 100C10

Hence the probability of getting at least one defective bulb is 1 – 80C10 / 100C10

(iv) none is defective

Solution:

Let M be the event that none of the selected bulb is defective. We know there are 80 good bulbs or 80 non- defective bulbs in the bag.

Hence number of outcomes of M= n{M}= 80C10

Since n{S}= 100C10, Probability of M= P{M}= n{M}/n{S}= 80C10 / 100C10

Hence the probability of getting all good bulbs is 80C10 / 100C10.

### Question 22. Find the probability that in a random arrangement of the letters of the word ‘SOCIAL’ vowels come together.

Solution:

We know there are 6 alphabets in the word ‘SOCIAL’

Hence the number of ways of arranging the letters = n{S}= 6!= 720

Let E be the event that vowels come together.

The vowels in SOCIAL are A, I, O. So, number of vowels= 3

Thus, number of ways to arrange them where the three vowels come together= n{E}= 4! × 3!= 144

Probability of E= P{E}= n{E}/n{S}= 144/720= 1/5.

Hence the probability of getting all the three vowels together is 1/5.

### Question 23. The letters of the word ‘CLIFTON’ are placed at random in a row. What is the chance that two vowels come together?

Solution:

We know there are 6 alphabets in the word ‘CLIFTON’

Hence the number of ways of arranging the letters = n{S}= 7!= 5040.

Let E be the event that vowels come together

The vowels in the word CLIFTON are I, O. So number of vowels= 3.

Thus, number of ways to arrange them where the three vowels come together= n{E}= 6! × 2!= 1440

Probability of E= P{E}= n{E}/n{S}= 1440/5040= 2/7.

Hence the probability of getting both vowels together is 2/7.

### Question 24. The letters of the word ‘FORTUNATE’ are placed at random in a row. What is the chance that two ‘T’ come together?

Solution:

We know there are 10 alphabets in the word ‘FORTUNATE’

Hence the number of ways of arranging the letters = n{S}= 10!

Let E be the event that two ‘T’ come together.

Thus, number of ways to arrange them where the three vowels come together= n{E}= 2 × 9!

Probability of E = P{E}= n{E}/n{S}= 2 × 9!/10!= 2/10= 1/5

Hence the probability of getting both ‘T’ together is 1/5.

### Question 25. A committee of two persons is selected from 2 men and 2 women. Find the probability that the committee will have:

(i) no man

Solution:

Since 2 people are to be selected from a group of 4 people, number of outcomes in sample space= n{S}= 4C2 = 6

Let M be the event denoting no man gets chosen from the committee. Hence, both the women get chosen.

Number of outcomes in event M= n{M}= 2C2 = 1

Probability of event M= P{M}= n{M}/n{S}= 1/6

Hence the probability of selecting no men is 1/6.

(ii) one man

Solution:

Let O be the event that one man is in the committee.

Since there are 2 men, number of outcomes in event O= n{S}= 2C1 × 2C1 = 2 × 2 = 4

Since n{S}= 6, Probability of event O=P{O}= n{O}/n{S}= 4/6

Hence the probability of choosing one man is 4/6.

(iii) two men

Solution:

Let T be the event that 2 men were chosen in the committee.

Since there are 2 men in the group, number of outcomes in event T= n{T}= 2C2 = 1

Since n{S}= 6, Probability of event T=P{T}= n{T}/n{S}=1/6

Hence the probability of choosing two men is 1/6.

### Question 26. If odds in favour of an event be 2:3, find the probability of occurrence of the event.

Solution:

Let the given event be denoted by E.

Since the odds are in the favour of the event are in the ratio 2:3, number of outcomes in sample space=n{S}= 2a + 3a= 5a.

Also, number of outcomes in event= n{E}= 2a

Probability of event E= P{E}= 2a/5a= 2/5

Hence the probability of occurrence of event E is 2/5.

### Question 27. If odds against an event be7:9, find the probability of non-occurrence of the event.

Solution:

Since the odds against the event are 7:9, number of outcomes in the sample space= n{S}= 7a + 9a= 16a.

Let E be the event that the particular event takes place.

Hence the number of outcomes in event E= 9a

Probability of E= P{E}= n{E}/n{S}= 9a/16a= 9/16

Thus, probability of non-occurrence= P{E’}= 1 − P{E}= 1 − 9/16= 7/16

Hence the probability of non-occurrence of event is 7/16.

### Question 28. Two balls are drawn at random from a bag containing 2 white, 3 red, 5 green and 4 black balls, one by one without replacement. Find the probability that both the balls are of different colors.

Solution:

The bag contains 2 + 3 + 5 + 4= 14 balls, and since two balls have been drawn without replacement, the total number of outcomes in the sample space= n{S}= 14C2 = 91.

Let E be the event that all the balls are of different colors and let E’ be the event that both the balls chosen are of the same color.

Event E’= {WW, RR, GG, BB}

Number of outcomes in event E’ = 2C2 + 3C2 + 5C2 + 4C2 = 20.

Probability of E’ = P{E’} = n{E’}/n{S} = 20/91

Thus, P{E} = 1 − P{E’}= 1 − 20/91= 71/91

Hence the probability of both the balls being of different colors is 71/91.

### Question 29. Two unbiased dice are thrown. Find the probability that:

(i) neither a doublet nor a total of 8 will appear

Solution:

Since two unbiased dice are thrown, the total number of outcomes in sample space= n{S}= 62 = 36.

Let E be the event that neither a doublet or a total of 8 will appear and E’ be the event that either a doublet or total of 8 will appear.

E’= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (2,6), (3,5), (5,3), (6,2)}

n{E’} = 10

Probability of E’= P{E’}= n{E’}/n{S}= 10/36

Thus, P{E}= 1 − P{E’}= 1 − 10/36= 26/36 = 13/18

Hence the probability that neither a doublet or a total of 8 will appear is 13/18.

(ii) the sum of the numbers obtained in the two dice is neither a multiple of 2 nor a multiple of 3.

Solution:

Let E be the event that the sum of the numbers obtained in the two dice is neither a multiple of 2 nor a multiple of 3 and E’ be the event that the sum of the numbers obtained in the two dice is either a multiple of 2 nor a multiple of 3.

E’= {(1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,5), (2,4), (3,3), (4,2), (5,1), (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (6,6)}

n(E’) = 24

Since n{S}= 36, Probability of E’= P{E’}= n{E’}/n{S}= 24/36

Thus, P{E}= 1 − P{E’}= 1 − 24/36= 12/36 = 1/3

Hence the probability that the sum of the numbers obtained in the two dice is neither a multiple of 2 nor a multiple of 3 is 1/3.

### Question 30. A bag contains 8 red, 3 white, and 9 blue balls. If three balls are drawn at random, determine the probability that

(i) all the three balls are blue balls.

Solution:

The bag contains 8 + 3 + 9 = 20 balls, and since three balls have been drawn at random, the total number of outcomes in the sample space= n{S} = 20C3 = 1140

Let E be the event that all the three balls are blue balls.

Number of outcomes in event E= 9C3 = 84.

Probability of E= P{E}= n{E}/n{S}= 84/1140 = 7/95

Hence the probability that all the three balls are blue balls is 7/95.

(ii) all the balls are of different colors.

Solution:

Let E be the event that all the balls are of different colors.

Number of outcomes in event E= 8C1 × 3C1 × 9C1 = 216.

Since n{S}= 1140, Probability of E= P{E}= n{E}/n{S}= 216/1140 = 18/95

Hence the probability that all the balls are of different colors is 18/95.

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