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

• Difficulty Level : Expert
• Last Updated : 19 Jan, 2021

### 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) = 52C4          -(1)

Let A be the event of getting cards of same color,

Since there are two sets of same color,

n(A) = 2 * 26C4           -(2)

P(A) = 2 * 26C4/52C4

= 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 theexaminations. 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)          -((A âˆ©  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) = 52C2          -(1)

Let A be the event of getting black cards,

n(A) = 26C2

P(A) =26C2/52C2           -(2)

Let B be the event of getting KING cards,

n(B) = 4C2

P(B) = 4C2/52C2           -(3)

Also, n(A âˆ© B) = 2C2

P(A âˆ© B) = 2/52C2          -(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) = 70C2          -(1)

Let A be the event of choosing the rusted item –

n(A) = 35C2.         -(20 nuts + 15 bolt = 35)

P(A) = 35C2/70C2

35 * 34/ 70 * 69          -(2)

Let B be the event of choosing two rusted bolts-

n(B) = 30C2

P(B) = 30C2/70C2

= 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 =16

Let 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)

(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

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) = 0.07

Suppose: A = {E1, E5, E8}, B ={E2, E5, E8, E9}          -(I)

Bc = {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(A âˆª B) = 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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