# Class 11 RD Sharma Solutions- Chapter 33 Probability – Exercise 33.4 | Set 2

### Question 15. From a pack of 52 cards, 4 cards are drawn at random. Find the probability that cards drawn are of the same color?

**Solution:**

From a pack of 52 cards, 4 cards are drawn

Hence, Sample space, n(S) =

^{52}C_{4}-(1)Let A be the event of getting cards of same color,

Since there are two sets of same color,

n(A) = 2 *

^{26}C_{4 }-(2)P(A) = 2 *

^{26}C_{4}/^{52}C_{4}= 92/833

_{ }Note: The factorial of the respective cases will give a large number,

so in such cases simplify the factorial in final step

**Question 16.** **100 students appeared for the two examinations, 60 passed the first, 50 passed the second and 30 passed both of the** **examinations. Find the probability that a student selected at random has at least passed one exam?**

**Solution:**

There are 100 students. Hence, sample space will be –

n(S) = 100 -(1)

Let A be the event that 60 students passed in first exam,

n(A) = 60

= 60/100 -(2)

Let B be the event that 50 students passed in first exam,

n(A) = 50

= 50/100 -(3)

30 passed both of the examinations,

P(A âˆ© B) = 30/100 -(4)

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 60/100 + 50/100 – 30/100 -(From 2, 3, 4)

= 4/5

### Question 17. A box contains 10 white, 6 red, and 10 black balls. A ball is drawn at random from the box, what is the probability that ball is red or white?

**Solution:**

There are 10 white, 6 red and 10 black balls, hence the sample space will be,

n(S) = 10 + 6 + 10

= 26

Let W be the event of drawing the White balls,

n(W) = 10

P(W) = 10/26 -(1)

Let R be the event of drawing the Red balls,

n(R) = 6

P(R) = 6/26 -(2)

E & R are mutually exclusive events –

n(R âˆ© E) = 0 -(

red and white balls can’t be drawn to be together)P(E âˆª R) = P(E) + P(R) – P(E âˆ© R)

= 10/26 + 6/26

= 16/26

= 8/13

### Question 18. In a race, the odds in favor of the horses A, B, C, D are 1 : 3, 1:4, 1:5 & 1:6. Find probability which one of them wins the race?

**Solution:**

We have P(A) : = 1 : 3

= P(A) / 1 – P(A) -(

= 1 – P(A))= 1/4 -(1)

Similarly P(B) = 1/5 -(2)

P(C) = 1/6 -(3)

P(D) = 1/7 -(4)

Probability that at least one of the horse wins is P(A âˆª B âˆª C âˆª D)

= 1/4 + 1/5 + 1/6 + 1/7 -(From 1, 2, 3, 4)

= 319/420

### Question 19. The probability that a person will travel by train is 3/5 and probability that he will travel by plane is 1/4. Find the probability that he will travel by train or plane ?

**Solution:**

Let T be the event that persons travels by train –

P(T) = 3/5 -(1)

Let A be the event that persons travels by PLANE –

P(A) = 1/4 -(2)

P(A âˆª T) = P(A) + P(T)

-((Aâˆ©T) = 0)= 3/5 + 1/4 -(From 1 , 2)

= 17/20

### Question 20. Two cards are drawn from a well-shuffled pack of 52 cards. Find the probability that both cards are black or king?

**Solution:**

Two cards are drawn from well shuffled deck of 52 cards,

n(S) =

^{52}C_{2 }-(1)Let A be the event of getting black cards,

n(A) =

^{26}C_{2}P(A) =

^{26}C_{2}/^{52}C_{2 }-(2)Let B be the event of getting KING cards,

_{ }n(B) =

^{4}C_{2}P(B) =

^{4}C_{2}/^{52}C_{2 }-(3)Also, n(A âˆ© B) =

^{2}C_{2}

_{ }P(A âˆ© B) = 2/^{52}C_{2 }-(4)Now,

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 26 * 25/52 * 51 + 4 * 3 / 52 * 51 – 2/52 * 51

= 55/221

### Question 21. In an entrance test graded on basis of two examination, the probability of randomly chosen student passing in exam 1 is 0.8 & the probability of passing in second examination is 0.7. The probability of passing in the one of the examination is 0.95. What is probability of passing in both?

**Solution:**

Let A be the event of selecting a random student who passed in examination 1,

P(A) = 0.8 -(1)

Let A be the event of selecting a random student who passed in examination 2,

P(B) = 0.7 -(2)

Now,

The probability of selecting a random student passing in at least one of the examination is –

P(A âˆª B) = 0.95 -(3)

Probability of passing both of the examination is P(A âˆ© B),

P(A âˆ© B) = P(A) + P(B) -P(A âˆª B)

= 0.8 + 0.7 – 0.95

=1.5-0.95

=0.55

### Question 22. A box contains 40 nuts & 30 bolts. Half of the bolts and nuts are rusted. If two items are drawn at random, what is probability that they are rusted and both are bolts?

**Solution:**

A box contains 40 nuts and 30 bolts,

half of them are rusted –

=> 40/2 = 20 nuts are rusted

=> 30/2 = 15 bolts are rusted

Since two items are drawn,

Sample space n(S) =

^{70}C_{2}-(1)Let A be the event of choosing the rusted item –

n(A) =

^{35}C_{2. }-(20 nuts + 15 bolt = 35)P(A) =

^{35}C_{2}/^{70}C_{2}35 * 34/ 70 * 69 -(2)

Let B be the event of choosing two rusted bolts-

n(B) =

^{30}C_{2}P(B) =

^{30}C_{2}/^{70}C_{2}= 30 * 29 / 70 * 69 -(3)

Also, n(A n B) = 15 -(

bolts are rusted)P(A n B) = 15*14/70* 69 -(4)

Now,

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= (35*34/70*69) + (30*29/70*69) – (15*14/70*69) -(From 2, 3, 4)

= 185/483

### Question 23. A integer is chosen at random from first 200 integers, find the probability that the integer is divisible by 6 or 8?

**Solution:**

A integer is chosen at random from 200 integers,

SAMPLE SPACE = n(S) = 200 -(1)

Let A be the event of choosing a number divisible by 6,

n(A) ={6,12,18. . . 198}

n(A) = 33 (

Using T n formula)P(A) = 33/200 -(2)

Let B be the event of choosing a number divisible by 8,

n(B) = {8,16,24. . . 200}

n(B) = 25

P(B) = 25/200

= 1/4 -(3)

Also, n(A n B)= {24,48. . 192}

= 8

P(A n B) = 8/200 -(4)

Now,

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 1/4 -(from 2, 3, 4)

### Question 24. Find the probability of getting 2 or 3 tails, when a coin is tossed 4 times?

**Solution:**

A coin is tossed 4 times,

Hence, Sample space – n(S) = 2

^{4 }=16Let A be the event of getting 2 tails,

A = {

HHTT, HTHT, TTHH, THTH, THHT, HTTH}n(A) = 6

P(A) = 6/16 -(1)

Let B the event of getting 3 tails,

B = {

HTTT, THTT, TTHT, TTTH}n(B) = 4

P(B) = 4/16 -(2)

A & B are the mutually exclusive events-

Hence, P(A âˆ© B) = 0

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 6/16 + 4/16-0

= 10/16

= 5/8

### Question 25. Suppose a integer is choose from 1 to 1000, find the probability that it is multiple of 2 or 9?

**Solution:**

A number is chosen from 1 to 1000,

Hence, sample space, n(S) = 1000

Number of multiples of 2 from 1 to 1000 are – 500

Number of multiples of 9 from 1 to 1000 are – 111

Out of 111, 55 are even numbers

Hence, total number multiple of 2 or 9 – 500 + 56

Probability of number multiple of 2 or 9 –

= 556/1000

= 0.556

### Question 26. In a metropolitan area the probabilities are 0.87, 0.36 & 0.30, that a family owns a color television set, a black and white TV or both. What is probability that family owns one of the set?

**Solution:**

Let A be the probability of a family having a color TV set,

P(A) = 0.87 -(1)

Let B be the probability of a family having a black and white TV set,

P(B) = 0.36 -(2)

Also, P(A âˆ© B) = 0.3

P(A âˆª B) = P(A) + P(B) – P(A âˆ© B)

= 0.87 + 0.36 – 0.30

= 0.93

### Question 27. If A and B are mutually exclusive events, such that P(A) = 0.35 and P(B) = 0.45 find,

**(i) P(A âˆª B)**

**(ii) P(A âˆ© B)**

**(iii)** [Tex] [/Tex]

**(iv) **** **

*** QuickLaTeX cannot compile formula: *** Error message: Error: Nothing to show, formula is empty

**Solution:**

(i)P(A âˆª B) = P(A) + P(B) (A and B are mutually exclusive events

i.e P( Aâˆ©B) = 0)= 0.80

(ii)P(A âˆ© B)P(A âˆ© B) = 0 (A and B are mutually exclusive events)

(iii)= P(A)= 0.35

(iv)= 1 – P(A âˆª B)= 1 – 0.80

= 0.20

### Question 28. A sample space consist of 9 elementary events, E1, E2, E3. . . E9, whose probabilities are P(E1) = P(E2) = 0.08, P(E3) = P(E4) = 0.1,P(E6) = P(E7) = 0.2,P(E8) = P(E9) = 0.07, suppose A = {E1, E5, E8}, B ={E2, E5, E8, E9}

**(i) compute P(A), P(B) & P(A âˆ© B)**

**(ii) Find P(A âˆª B)**

**(iii) List the combustion of the event A âˆª B, and find P(A âˆª B)**

**(iv) Calculate **

*** QuickLaTeX cannot compile formula: *** Error message: Error: Nothing to show, formula is empty

** **

**Solution:**

Given:P(E1) = P(E2) = 0.08, P(E3) = P(E4) = 0.1, P(E6) = P(E7) = 0.2, P(E8) = P(E9) = 0.07Suppose: A = {E1, E5, E8}, B ={E2, E5, E8, E9} -(I)

B^{c}= {E1,E3,E4,E6.E7}

P(E5) =1 – (0.08 + 0.1 + 0.2 + 0.07}=0.1

(i)P(A) = 0.08 + 0.1 + 0.07 (From Given)= 0.25

P(B) = 0.08 + 0.1 + 0.07 + 0.07

= 0.32

P(A âˆ© B) = 0.1 + 0.07

= 0.17

(ii)P(AâˆªB) = P(A) + P(B) – P(A âˆ© B)= 0.57-0.17

= 0.40

(iii)= 1 – P(B)= 1 -0.32

= 0.68

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