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

**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}=

^{16}C_{2}= 120.Let W be the event of getting both white balls.

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

^{7}C_{2}= 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}=

^{5}C_{1}Ã—^{ 4}C_{1}= 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}=

^{7}C_{2}+^{5}C_{2 }+^{4}C_{2}= 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}=

^{18}C_{3}= 816.Let Y be the event of getting one red and two white balls.

Hence number of outcomes in Y = n{Y} =

^{6}C_{1}Ã—^{4}C_{2 }= 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}=

^{8}C_{2}Ã—^{6}C_{1}=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}=

^{6}C_{1}Ã—^{4}C_{1}Ã—^{8}C_{1}+^{6}C_{1}Ã—^{4}C_{2}+^{6}C_{1}Ã—^{8}C_{2}= 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}=

^{52}C_{5}= 2598960Let E be the event that exactly only one ace is present.

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

^{4}C_{1}Ã—^{ 48}C_{4 }= 778320Probability 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} =

^{4}C_{1}Ã—^{48}C_{4}+^{4}C_{2}Ã—^{48}C_{3}+^{4}C_{3}Ã—^{48}C_{2}+^{4}C_{4}Ã—^{48}C_{1 }= 886656Probability 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}=

^{40}C_{4}= 91390Let E be the event that 4 cards belong to different suit.

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

^{10}C_{1}Ã—^{10}C_{1}Ã—^{10}C_{1}Ã—^{10}C_{1}= 10000Probability 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}=

^{10}C_{1 }= 10.Let E be the event that it is a woman.

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

^{6}C_{1}= 6Probability 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}=

^{100}C_{10}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}=

^{20}C_{10}Probability of event E= P{E} = n{E}/n{S}=

^{20}C_{10 }/^{100}C_{10}

Hence the probability of all 10 bulbs being defective is^{20}C_{10 }/^{100}C_{10}.

**(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= 100âˆ’ 20= 80.

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

^{80}C_{10}Since n{S}=

^{100}C_{10}, Probability of event U=P{U}= n{U}/n{S}=^{80}C_{10 }/^{100}C_{10}

Hence the probability of all balls being good is^{80}C_{10 }/^{100}C_{10}.

**(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 Râ€² be the event that none of the bulb is defective

n{R’}=

^{80}C_{10}Since n{S}=

^{100}C_{10}, Probability of R’= P{R’}= n{R’}/n{S}=^{80}C_{10 }/^{100}C_{10}So, P{R}= 1 â€“ P{R’}= 1 â€“

^{80}C_{10}/^{100}C_{10}

Hence the probability of getting at least one defective bulb is 1 â€“^{80}C_{10}/^{100}C_{10}

**(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}=

^{80}C_{10}Since n{S}=

^{100}C_{10}, Probability of M= P{M}= n{M}/n{S}=^{80}C_{10}/^{100}C_{10}

Hence the probability of getting all good bulbs is^{80}C_{10}/^{100}C_{10}.

**Question 22. Find the probability that in a random arrangement of the letters of the word â€˜SOCIALâ€™ 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! Ã— 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 â€˜CLIFTONâ€™ 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! Ã— 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 â€˜FORTUNATEâ€™ 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 ‘FORTUNATEâ€™

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 Ã— 9!

Probability of E = P{E}= n{E}/n{S}= 2 Ã— 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}=

^{4}C_{2}= 6Let 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}=

^{2}C_{2}= 1Probability 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}=

^{2}C_{1}Ã—^{2}C_{1}= 2 Ã— 2 = 4Since 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}=

^{2}C_{2}= 1Since 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 âˆ’ P{E}= 1 âˆ’ 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}=

^{14}C_{2}= 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’ =

^{2}C_{2}+^{3}C_{2}+^{ 5}C_{2}+^{4}C_{2 }= 20.Probability of E’ = P{E’} = n{E’}/n{S} = 20/91

Thus, P{E} = 1 âˆ’ P{E’}= 1 âˆ’ 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 âˆ’ P{E’}= 1 âˆ’ 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 âˆ’ P{E’}= 1 âˆ’ 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} =

^{20}C_{3}= 1140Let E be the event that all the three balls are blue balls.

Number of outcomes in event E=

^{9}C_{3}= 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=

^{8}C_{1}Ã—^{3}C_{1}Ã—^{9}C_{1}= 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.

## Please

Loginto comment...