## Rajasthan Board RBSE Class 11 Maths Chapter 14 Probability Miscellaneous Exercise

Question 1.

In throwing a coin n times, n(S) is –

(A) 2n

(B) 2^{n}

(C) n^{2}

(D) n/2

Solution:

Thus (B) is correct.

Question 2.

In throwing two dice and sample space of getting sum 3 is –

(A) (1, 2)

(B) {(2, 1)}

(C) {(3, 3)}

(D) {(1, 2), (2, 1)}

Solution:

Thus (D) is correct.

Question 3.

If tossing a coin and a dice simultaneously the number of elements in sample space is—

(A) 12

(B) 6

(C) 64

(D) 36

Solution:

No. of elements = n{H, T} × n{1, 2,3,4,5,6}

= 2 × 6 = 12

Thus, (A) is correct.

Question 4.

Result of each experiment is called –

(A) Sample space

(B) Random test

(C) Sample point

(D) Ordered pair

Solution:

Thus, (C) is correct.

Question 5.

If three coin are tossed and E be the event to getting at least one head, then n(E) will be –

(A) 6

(B) 3

(C) 4

(D) 8

Solution:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, ITT)

n(E) = 4

Thus, (C) is correct.

Question 6.

if E_{1} ∩ E_{2} = Φ, then E_{1} and E_{2} will be –

(A) Exclusive

(B) Independent

(C) Dependent

(D) Complementary

Solution:

Thus, (D) is correct.

Question 7.

Favourate events of 53 Mondays in leap year will be –

(A) 7

(B) 2

(C) 1

(D) 14

Solution:

Total days in a leap year = 366

Remaining day of after 53 weeks

= \(\frac { 366 }{ 53 } \) = 2 remaining

Thus, (B) is correct.

Question 8.

Three are 4 white 5 black and 2 red balls in an urn. Favourable cases of three different colour balls will be –

(A) 9

(B) 24

(C) 12

(D) 7

Solution:

White balls = 4, Black balls = 3

and Red balls = 2

Favourable cases of three different colour balls

= 4 × 3 × 2 = 24

Thus, (B) is correct.

Question 9.

In two mutually exclusive events value of P (A ∪ B) is –

(A) P(A) + P(B)

(B) P(A) + P(B) – P(A ∩ B)

(C) P(A).P(B)

(D) P(A).P(B/A)

Solution:

Thus, (A) is correct.

Question 10.

The probability of solving the question by three students A,B and C are 1/2, 1/3 and 1/4, then probability of solving the question by at least one is –

(A) 1/24

(B) 1/4

(C) 3/4

(D) 1/9

Solution:

P(A)= 1/2, P(B) = 1/3, P(C) = 1/4

Required probability

Thus, option (C) is correct.

Question 11.

On tossing two dices simultaneously, probability to getting diference of numbers appear as 1 will be –

(A) 5/18

(B) 1/4

(C) 2/9

(D) 7/36

Solution:

Total events = 6 × 6 = 36 ⇒ n(S) = 36

Favourable events

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

n(A) = 10

Thus, Required probability

Thus, (A) is correct.

Question 12.

A card is drawn from a deck of cards, probability of getting red or black card is –

(A) 1/4

(B) 1/2

(C) 3/4

(D) 26/51

Solution:

Required probability

Thus, (B) is correct.

Question 13.

On throwing two dices probability to getting sum of numbers appear as multiple of 4 will be –

(A) 1/4

(B) 1/3

(C) 1/9

(D) 5/9

Solution:

Multiples of 4 are = 4, 8, 12, 16,20,24, 28, 32, 36, 40, … etc.

Thus favourable events A = {(1, 3), (3, 1), (4,4), (5, 3), (3,5), (6, 2), (6,6)}

n(A) = 9 and n(S) = 36

Thus, required probability

Thus, (A) is correct.

Question 14.

If 5 digit numbers are formed by using digit 1,2,3,4, 5,6 and 8, then probability to get even digit at both end will be –

(A) 5/7

(B) 4/7

(C) 3/7

(D) 2/7

Solution:

No. of total digits = 7

Even number appear on both end so, favourable cases = 2

Thus, required probability = \(\frac { 2 }{ 7 } \)

Thus, (D) is correct.

Question 15.

In throwing three dices probability to get same digit on all three is –

(A) 1/36

(B) 3/22

(C) 1/6

(D) 1/18

Solution:

∵ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}

n(A) = 6, n(S) = 6^{3} = 216

Required probability

Thus, (A) is correct.

Question 16.

In a swimming race odds in favour of A is 2 : 3 and odds in opposite of B is 4 : 1. Find the probability of winning of A orB is –

(A) 1/5

(B) 2/5

(C) 3/5

(D) 4/5

Solution:

Question 17.

10 students are sitting in a raw randomly, probability that two specific students do not sit closely is –

(A) 1/5

(B) 2/5

(C) 3/5

(D) 4/5

Solution:

Required probability

Thus, (D) is correct.

Question 18.

There are 12 sections in a group, in which four sections are faulty 3 sections are randomly drawn one by one. Without replacement probability to get no. one as faulty is –

(A) 3/55

(B) 13/55

(C) 14/55

(D) 17/55

Solution :

Total terms = 12, Non-facility sections = 12 -4 = 8

Faulty section = 4

Probability to get no. one as fault in three attempt is –

Thus (C) is correct.

Question 19.

Probability of any sure event will be –

(A) 0

(B) 1/2

(C) 1

(D) 2

Solution:

Thus (C) is correct.

Question 20.

A family having 3 children in which at least one boy, then probability that family have one boy and one girl is –

(A) 1/2

(B) 1/3

(C) 1/4

(D) 3/4

Solution:

Sample space of children in a family is :

S = {B, G, G, B, B, G, B, B, G}

Event of 2 boys and 1 girl A = {B, B, G}

∴ n(S) = 3, n(A) = 1

Thus, required probability = \(\frac { n(A) }{ n(S) } \) = \(\frac { 1 }{ 3 } \)

Thus, (B) is correct.

Question 21.

The probability of taking examination in class by a teacher is 1/5. If a student is absent two times then probability that he was absent at least in one examination is –

(A) 9/25

(B) 11/25

(C) 13/25

(D) 23/25

Solution:

Let student is absent in first examination, event is collect E and absent in 2nd is F, then

According to question

Probability that students is absent in at least one exam. =

1 – (Prob. that student is absent in both the exams)

Thus, this is prob. that student appear at least one examination.

Thus, (A) is correct.

Question 22.

In a non-leap year. Find the probability to getting 53 Mondays.

Solution:

There are 52 Sundays in a leap year and 2 days are remains. If one of them is Sunday, then there will be 53 Sundays.

Now, probability that at least two days one of them is Sunday = \(\frac { 1 }{ 7 } \)

∴ Required probability = \(\frac { 1 }{ 7 } \)

Question 23.

A and B are two mutually exclusive events and P(A) – 0.3, P(B) = K and P(A ∪B) = 0.5, then find the value of K –

Solution:

P(A ∪ B) = P(A) + P(B)

⇒ P{B) = P(A ∪ B) – P(A)

⇒ P(B) = K = 0.5 – 0.3

= 0.2

Thus K = 0.2.

Question 24.

Words are formed by using letters of word ‘PEACE’. Find the probability that word contains both E.

Solution:

Total words formed by letters of word ‘PEACE’

Question 25.

There are 6 red and 8 black balls in a bag. 4 balls are taken out two times, 4 balls taken once are replaced back. What will be the probability that 4 balls in first attempt be red and in second attempt be black ?

Solution:

Total balls = 6 + 8 = 14

Since ball are replaced

∴ Probability to draw 4 red ball in 1st chance and 4 black balls in 2nd chance.

Question 26.

A man speaks truth 3 times out of 5. He says that in tossing 6 coins,two tails appear so what is the probability that this event is actually true ?

Solution:

Let E represents the statement of a person. Now, S_{1} is the event of getting 2 heads and S_{2} is the event of not getting 2 heads on tossing 6 coins

Now, Probability of statement to be truth if 2 heads are obtained on tossing of 6 coins

∵ Person speaks truth 3 times out of 5.

Similarly, Probability of statement to be not truth if 2 heads are obtained on tossing of 6 coins

Now, the probability that this event is actually true,

Question 27.

In throwing two dices, what is the probability that neither same digit appear nor sum of digit be 9 ?

Solution:

Total events in throwing two dices simultaneously

n(S) = 6^{2} = 36

Total events to get same digit at two dices and sum 9

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

n(A) = 10

Probability to get same digit on dices or sum 9

Thus, required probability

Question 28.

Three coins are tossed simultaneously, then find the probability where as –

(i) Exactly two heads appear.

(ii) At least two heads appear,

(iii) Maximum two head appear.

(iv) AH the three are heads.

Solution:

In throwing three coins obtained sample space

S = {(HHH), (HHT), (7/77/7), (HTT), (THH), (THT), (TTH), (TTT)}

n(s) = 8

(i) Probability to get exactly two heads = \(\frac { 3 }{ 8 } \)

(ii) Events if at least two heads apear

A = {(HHH), (HHT), (HTH), (THH)}

∴ n(A) = 4

Thus, required probability

(iii) Event to get maximum 2 heads –

A = {(HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}

n(A) = 7

Thus, required probability

(iv) Events to get all the three heads

A = {HHH}

n(A) = 1

Thus, required probability

Question 29.

In a horse race, four horses A, B, C, D run odds in favour of A, B, C, D are respective of 1 : 3, 1 : 4, 1: 5 and 1: 6. Find the probability that are one of them wins.

Solution:

Let event to win horses A, B, C, D are E, F, G, H respectively.

Since only one is the winner

Thus, these are mutually exclusive events

∴ Required probability

Question 30.

Probability that a person will alive in next 25 years is 3/5 and of his wife 2/3. Find the following probability:

(i) Both remains alive.

(ii) None of two remains alive.

(iii) At least one remains alive.

(iv) Only wife remains alive.

Solution:

According to question, prob. of one person alive

(i) Prob. that both remains alive

By formula P(AB) = P(A).P(B)

P(EF) = P(E).P(F)

= \(\frac { 3 }{ 5 } \) × \(\frac { 2 }{ 3 } \) = \(\frac { 2 }{ 5 } \)

(ii) Probability that none of them remain alive

(iii) Probability that at least one remain alive

(iv) Prob. that only wife remain alive

Question 31.

A and B are independent witness. Probability that A speaks truth is x and y of B if A agree with B for any statement, then prove that probability of the truth of statement will be xy/(1 – x + 2xy).

Solution:

5 The prob. that A speaks truth

P(A) = x

Probability that B speaks truth

P(B) = y

If both are agree with any statement then, prob. that statement is true

Question 32.

Three males A,B C tossed a coin one by one. Who gets tail first he will win. If A has first chance, then what is the probability to winning of A.

Solution:

Prob. to get tail = \(\frac { 1 }{ 2 } \)

and Pro. to not get tail = \(\frac { 1 }{ 2 } \)

∵ A toss a coin in first chance thus he may win 1st time, 4th time, 7th time.

∴ Prob. that A win

Similarly prob. that B wins

Similarly prob. that C wins

Question 33.

Sulakshina and Sunayana tossed a coin one by one. Who get tail first she will win.If Sulakshina has first chance, then find probability that both of them win.

Solution:

Prob. to get.tail = \(\frac { 1 }{ 2 } \)

and prob. to not get tail = \(\frac { 1 }{ 2 } \)

∵ Since first, sulakshina toss a coin

Thus, she may win in 1st, 3rd, 5th attempt

∴ Probability that Sulakshina wins

and probability that sunayana wins

Question 34.

One digit is selected from the following two groups of numbers –

(1, 2, 3, 4, 5, 6, 7, 8, 9), (1, 2, 3, 4, 5, 6, 7, 8, 9)

If P_{1} is sum of both digit as 10, P_{2} is sum of both digit as 8, then find P_{1} + P_{2}.

Solution:

P_{1} = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

n(P_{1}) = 9

Total ways = 9 × 9 = 81 = n(S)

Question 35.

If P(A) = 0.4, P(B) = 0.8, P(B/A) = 0.6, then find P(A/B) and P(A ∪ B).

Solution:

By formula

Question 36.

If P(E) = 0.35, P(P) = 0.45, P(E ∪F) = 0.65, then find P(F/E).

Solution:

Question 37.

A dice is thrown five times, find the probability getting only one.

Solution:

Probability to get 1 in throwing a dice = (\(\frac { 1 }{ 6 } \))

and Probability not get = 1 – \(\frac { 1 }{ 6 } \) = \(\frac { 5 }{ 6 } \)

Thus probability to get only digit 1 in five tosses

## Leave a Reply