Probability MCQS

Question 1.
A letter is taken out at random from the word RANGE and another is taken out from the word PAGE. The probability that they are the same letters is : 
(a) 1/20
(b) 3/20
(c) 3/5
(d) 3/4
Solution:
(b) 3/20

Letters A ; G ; E are common.
Total probability for each Letter in both words
= 15 × 14 + 15 × 14 + 15 × 14 = 320

Question 2.
An urn contains 9 balls two of which are red, three blue and four black. Three balls are drawn at random. The probability that they are of same colour is:
(a) 327
(b) 2031
(c) 584
(d) None
Solution:
(c) 584 = 84 ways
E = Event same colour
n(S) = ³C₃ + ⁴C₃ = 1 + 4 = 5
(Since, either all the balls can be blue or black but not red)
P(E) = 584

Question 3.
A card is drawn from a well shuffled pack of 52 cards. Let E, “a king or a queen is drawn” & E₂: “a queen or a jack is drawn”, then:
(a) E₁ and E₂ are not independent
(b) E₁ and E₂ are mutually exclusive
(c) E₁ and E₂ are independent
(d) None of these [1 Mark, Nov. 2006]
Solution:
(a) E₁ and E₂ are not independent
Note only one experiment made, so, they are not Independent.
Detail:- E₁ = a king or a queen is drawn
E₂ = a queen or a jack is drawn
P(E₁) = 4+452 = 852
P(E₂) = 4+452 = 852
P(E₁ ∩ E₂) = p (drawing a queen) = 452
Here P(E₁ ∩ E₂) ≠ P(E₁).P(E₂)
Hence E₁ and E₂ are not independent

Question 4.
In a non-leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is : 
(a) 47
(b) 27
(c) 37
(d) 17
Solution:
(c) 37

1 non – leap year = 365 days,
= 52 complete weeks and 1 additional day
S = (Sun.; Mon.; Tue.; Wed.; Thur.; Fri.;Sat.}; n (S) = 7
Required Event = E = {Sun.,Tue.,Thur.}
n (E) = 3
P(E) = 37

Question 5.
If A andB are two events and P(A) = 38, P(B) = 12, P(A∩B) = 14, then the value of P(A’ ∪ B’) is 
(a) 14
(b) 34
(c) 58
(d) 54
Solution:
(b) 34
P(A ∩ B) = 14
According to De-Morgans Law
P(A’ ∪ B’) = P(A∩B)’
= 1 – P(A∩B) = 1 – 14 = 34

Question 6.
From a pack of cards, two are drawn, the first being replaced before the second is drawn. The chance that the first is a diamond and the second is king is: 
(a) 152
(b) 32704
(c) 413
(d) 352
Solution:
(a) 152
Required probability
= P(card drawn is a diamond) × P (card drawn is a king)
= 1352 × 452(since there are 13 diamonds and 4 kings).
= 152

Question 7.
The theory of compound probability states that for any two events A and B:
(a) P (A ∩ B) = P(A) × P(B)
(b) P (A ∩ B) = P(A) × P(B/A)
(c) P (A ∪ B) = P(A) × P(B/A)
(d) P (A ∪ B) = P(A) + P(B) – P(A ∩ B) [1 Mark, May 2007]
Solution:
(b) P (A ∩ B) = P(A) × P(B/A)
Simultaneous Occurrence of two events A and B are Compound Events
For it; Formula
P(A∩B) = P(A) × P(B/A)

Question 8.
The probability of getting qualified in IIT – JEE and AIEEE by a student are respectively 15 and 35. The probability that the student gets qualified for one of the these tests is :
[1 Mark, May 2007]
Solution:
(a) 1725
(b) 2225
(c) 825
(d) 325
Solution:
(a) The Events for being selected for IIT JEE & AIEEE are E₁ & E₂ respectively.
So, P(E₁) = 15; P(E₂) = 35
The probability that he is selected for one of these test = 1 – P (selection in no tests)
= 1 – P(E₁‘)P(E₂‘)
= 1 – (1 – 15) × (1 – 35) = 1725

Question 9.
Suppose E and F are two events of a random experiment. If the probability of occurrence of E is 15 and the probability of occurrence of F given E is 110, then the probability of non-occurrence of at least one of the events E and F is :
(a) 150
(b) 125
(c) 1350
(d) 4950
Solution:
(d) 4950
Given P(E) = 15, P(F/E) = 110
∴ P(E ∩ F) = P(E).P(F/E)
= 15.110 = 150
∴ P(non-occurrence of at least one of the events of E and F)
P(E’∪F’) = P(E∩F)’
= 1 – P(E ∩ F)
= 1 – 150 = 4950

Question 10.
A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made without replacement. The probability that the first draw will produce 3 white balls and second 3 red balls is : 
(a) 6255
(b) 5548
(c) 7429
(d) 3233
Solution:
(c) 7429
P(1st 3 white then 3 red balls)
= P (White). P (Red) =
5C313C3 . 8C310C3 = 10286 × 56120 = 5143 × 715 = 7429

Question 11.
Three identical dice are rolled. The probability that the same number will appear on each of them is: 
(a) 1/6
(b) 1/12
(c) 1/36
(d) 1
Solution:
(c) 1/36
n (S) = 6³ = 216
Event A = same number appears on each of the three dice.
= {(1, 1, 1) ; (2, 2, 2) ; (3, 3, 3) ; (4, 4, 4) ; (5, 5, 5) ; (6, 6, 6)}
So, n(A) = 6
P(E) = 66×6×6 = 136

Question 12.
Among the examinees in an examination 30%, 35% and 45% failed in Statistics, in Mathematics and in at least one of the subjects respectively. An examinee is selected at random. Find the probability that he failed in Mathematics only: 
(a) 0.245
(b) 0.25
(c) 0.254
(d) 0.55
Solution:
(a) 0.245
Event S = Students failed in Statistics, and
Event M = students failed in Mathematics
P(S) = 0.30,
P(M) = 0.35
and ∴ P(S ∪ M) = 0.45
Hence P(failed in mathematics) = P (failed in Mathematics but passed in Statistic)
= P(M) × P(S’)
= 0.35 × (1 – 0.30) = 0.245

Question 13.
An article consists of two parts A and B. The manufacturing process of each part is such that probability of defect in A is 0.08 and that in B is 0.05. What is the probability that the assembled product will not have any defect? 
(a) 0.934
(b) 0.864
(c) 0.85
(d) 0.874
Solution:
(d) 0.874

Let X = Event (Part A is non-defective) and Y = Event (Part B is non-defective)
∴ P(X) = 1 – 0.08 = 0.92
P(Y) = 1 – 0.05 = 0.95
P(Products are non-defective) = P(X ∩ Y) = P(X).P(Y) = 0.92 × 0.95 = 0.874

Question 14.
If 10 men, among whom are A and B, stand in a row, what is the probability that there will be exactly 3 men between A and B ?
(a) 11/15
(b) 4/15
(c) 1/15
(d) 2/15
Solution:
(d) 2/15
There are 2 men standing in a row.
A and B are 2 men can take any 2 positions out of 10 in ¹⁰C₂ = 45 ways.
So, n(S) = 45.(2!) = 45 × 2 = 90. (∵ A and B can interchange their positions)
According to the Qts.
E = Event of placing 3 men between A & B = {(1, 5) ;(2, 6);(3, 7);(4, 8); (5, 9); (6 ,10); (5, 1), (6, 2), ; (10, 6)}
n(E) = 6.(2!) = 6 × 2 = 12
P(E) = n(E)n(S) = 6×245× = 215

Question 15.
The probability of an event can assume any value between: 
(a) 0 and 1
(b) -1 and 0
(c) -1 and 1
(d) None of these
Solution:
(d) None of these
Since the Probability of an event lies between 0 and 1, both inclusive i.e. 0 ≤ P(A) ≤ 1.

Question 16.
The odds are 9 : 5 against a person who is 50 years living till he is 70 and 8 : 6 against a person who is 60 living till he is 80. Find the probability that at least one of them will be alive after 20 years:
(A) 1114
(B) 2249
(C) 3149
(D) 3549
Solution:
(C) 3149
Event A : Person aged 50 years will remain alive for next 20 years
Event B: Person aged 60 years will remain alive next 20 years
Given: P(A′)P(A) = 95 & P(B′)P(B) = 86
∴ P(A’) = 99+5 = 914 & P(B’) = 88+6 = 814
P (Atleast one will remain alive) = 1 – P (None will remain alive)
= 1 – P(A’).P(B’) = 1 – 914.814 = 124196 = 3149

Question 17.
If P(A) = P and P (B) = q, then:
(a) P (A/B) ≤ q/p
(b) P(A/B) ≥ p/q
(c) P(A/B) ≤ p/q
(d) P(A/B) ≥ q/p 
Solution:
(c) P(A/B) ≤ p/q
Given P(A) = P and P(B) = q
We know that
P(A/B) = P(A∩B)P(B)
and also
P(A ∩ B) < P(A)
Dividing by P(B) on both sides ; We get
P(A∩B)P(B) ≤ P(A)P(B)
∴ P(A/B) ≤ p/q

Question 18.
The probability that a trainee will remain with a company is 0.8. The probability that an employee earns more than ₹ 20,000 per month is 0.4. The probability that an employee, who was a trainee and remained with the company or who earns more than ₹ : 20,000 per month is 0.9. What is the probability that an employee earns more than ₹ 20,000 per month given that he is a trainee, who stayed with the company ?
Solution:
(a) 5/8
(b) 3/8
(c) 1/8
(d) 7/8
Solution:
(b) 3/8

Let Event A = A trainee will remain with the company & Event B = An Employee earns more than ₹ 20,000.
Given P(A) = 0.8, P(B) = 0.4 and P(A ∪ B) = 0.9
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.9 = 0.8 + 0.4 – P(A ∩ B)
⇒ P(A ∩ B) = 0.3
Hence P(B/A) = P(B∩A)P(A) = 0.30.8 = 38

Question 19.
The limiting relative frequency of probability is: 
(a) Axiomatic
(b) classical
(c) statistical
(d) Mathematical
Solution:
(c) statistical

The limitations of the statistical definition of probability, based on the concept of relative frequency:—
Let an event A occurs A times. Then, the limiting value of the ratio of f_A to n as n tends to infinity is defined as the probability of A.
Therefore, the limiting relative frequency of probability is Statistical.

Question 20.
If A and B are two independent events and P (A ∪ B) = 2/5; P(B) = 1/3 Find P(A)
(a) 2/9
(b) -1/3
(c) 2/10
(d) 1/10 
Solution:
(d) 1/10
Since A and B are two independent events, therefore –
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A). P(B) = P(A) + P(B) – P(A ∪ B)
⇒ P(A)(13 – 1) = 5−615
⇒ P(A).(−23) = −115
∴ P(A) = 110

Question 21.
A bag contains 12 balls of which 3 are red 5 balls are drawn at random Find the probability that in 5 balls 3 are red. 
(a) 3/132
(b) 5/396
(c) 1/36
(d) 1/22
Solution:
(d) 1/22
Sample Space n (S) = ¹²C₅
No. of Red balls = 3 ;
Other colour balls = 9;
n (E) = Event of 3 red balls if 5 balls are drawn. = ³C₃ . ⁹C₂
P(E) = 3C3×9C212C5 = 1×36792 = 122

Question 22.
P(A) = 2/3; P(B) = 3/5; P(A∪B) = 5/6. Find P(B/A)
(a) 11/20
(b) 13/20
(c) 13/18
Solution:
(b) 13/20
∵ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 23 + 35 – 56
= 20+18−2530 = 1330

Question 23.
If P (A ∩ B) = P (A) × P (B), then the events are:
(a) Independent events
(b) Mutually exclusive events
(c) Exhaustive events
(d) Mutually inclusive events
Solution:
(a) Independent events
If A and B are two independent events, then P(A ∩ B) = P(A) × P(B).

Question 24.
In a pack of playing cards with two jokers probability of getting king of spade is
(a) 4/13
(b) 4/52
(c) 1/52
(d) 1/54
Solution:
(d) 1/54

Total No. of playing cards = 52 + 2 = 54
sample space n(S) = ⁵⁴C₁ = 54
Total no. of spade king = 1
E = Event of getting 1 spade king = 1
∴ P(E) = 154

Question 25.
Consider two events A and B not mutually exclusive, such that P(A) = 1/4, P(B) = 2/5, P(A ∪ B) = 1/2, then P(AB¯¯¯¯) is 
(a) 3/7
(b) 2/10
(c) 1/10
(d) None of them
Solution:
(d) None of them
Since events A & B are not mutually exclusive. So, they are independent events.
P(AB) = P(A ∩ B) = P(A) .P(B)
∴ P(A B¯¯¯¯) = P(A) – P(A ∩ B) = P(A) – P(A).P(B)
= P(A).[1 – (B)]
= 14.(1 – 25) = 14 . 35 = 320

Question 26.
If x be the sum of two numbers obtained when two dice are thrown simultaneously then P(x ≥ 7) is
(a) 5/12
(b) 7/12
(c) 11/15
(d) 3/8
Solution:
(b) 7/12
Sample Space n (S) = 36
E = Event of getting X > = 7
= {(1, 6) ; (2, 5) ; (3, 4) ; (4 ; 3) ; (5, 2) ; (6, 1); (2, 6) ; (3, 5) ; (4, 4) ; (5, 3) ; (6, 2); ………;(6,6)}; ∴ n(E) = 21
∴ Required Probability P(E) = 2136 = 712
Tricks See Nov. 2019 Qts.

Question 27.
If P(A/B) = P(A), then A and B are
(a) Mutually exclusive events
(b) Dependent events
(c) Independent events
(d) Composite events 
Solution:
(c) Independent events
P(A/B) = P(A)
Hence, A and B are independent events.

Question 28.
A bag contains 3 white and 5 black balls and second bag contains 4 white and 2 black balls. If one ball is taken from each bag, the probability that both the balls are white is _____ 
(a) 1/3
(b) 1/4
(c) 1/2
(d) None of these
Solution:
(b) 1/4

Let A = Event of getting 1 white ball from 1st bag.
B = Event of getting 1 white ball from 2nd bag.
A & B are Independent Events.
P(A ∩ B) = P(A).P(B)
3C18C1⋅4C16C1 = 38×46 = 14

Question 29.
The odds in favour of A solving a problem is 5:7 and Odds against B solving the same problem is 9:6. What is the probability that if both of them try, the problem will be solved?
(a) 117/180
(b) 181/200
(c) 147/180
(d) 119/180 
Solution:
(a) 117/180

P(A) = 512 ; P(B) = 615
Probability that problem will be Solved = 1 – Prob. that problem not solved
= 1 – P(A’).P(B’)
= 1 – [1 – P(A)][1 – P(B)]
= 1 – (1 – 512)(1 – 615)
= 1 – 712 × 915 = 180−63180 = 117180

Question 30.
Consider Urn 1:2 white balls, 3 black balls; Urn II: 4 white balls, 6 black balls. One ball is randomly transferred from first to second Urn, then one ball is drawn from II Urn. The probability that drawn ball is white is
(a) 22/65
(b) 22/46
(c) 22/55
(d) 21/45 
Solution:
(c) 22/55

Case I: 1 Black ball is transferred from Urn I to Urn II and then a white ball is picked from Urn II
Prob. = 35 × 411 = 1255
Case II: 1 White ball is transferred from Urn I to II and then a white ball is picked from Urn II
Prob. = 25 × 511 = 1055
From Case I + Case II;
P(Whiteball) = 1255 + 1055 = 2255

Question 31.
If P (A∪B)= P(A), Find P(A∩B).
(a) P(A).P(B)
(b) P(A) + P(B)
(c) 0
(d) P(B) 
Solution:
(d) P(B)
Given : P(A ∪ B) = P(A) we know that
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ we get P(A ∩ B) = P(B)

Question 32.
A bag contains 5 Red balls, 4 Blue Balls and ‘m’ Green Balls. If the random probability of picking two green balls is 1/7. What is the No. of green Balls (m)
(a) 5
(b) 7
(c) 6
(d) None of above 
Solution:
(c) 6
Total balls = 5 + 4 + m = m + 9
∴ Probability of picking two green balls = 17

TRICKS : Go by choices
For (c) Total = 6 + 9 = 15
P = 6c215c2
= 15105 = 17 (True)

Question 33.
The probability of Girl getting scholarship is 0.6 and the same probability for Boy is 0.8 Find the probability that at least one of the categories getting scholarship. 
(a) 0.32
(b) 0.44
(c) 0.92
(d) None of these
Solution:
(c) 0.92

Given
Probability of Girl getting scholarship P(A) = 0.6
Probability of Boy getting scholarship P(B) = 0.8
Required of at least one category getting scholarship:
= 1 – P(None getting scholarship)
= 1 – P (A’) .P(B’) = 1 – (1 – 0.6).( 1 – 0.8)
= 1 – (0.4). (0.2) = 1 – 0.08 = 0.92

Question 34.
A coin is tossed 5 times, what is the probability that exactly 3 heads will occur.
(a) 516
(b) 132
(c) 536
(d) 332
Solution:
(a) 516

(a) Total No. of Tails (n) = 5
r = 3
Probability of getting Head (p) = 1/2
Probability of getting Tail (q) = 1/2
By Binomial Distribution,
P(x = r) = ⁿC_rp^rq^n-r
P(r = 3) = ⁵C₃(12)3(12)5−3 = 10.125 = 1032 = 516

Question 35.
Arun & Tarun appear for an interview for two vacancies. The probability of Arun’s selection is 1/3 and that of Tarun’s selection is 1/5. Find the probability that only one of them will be selected.
(a) 2/5
(b) 4/5
(c) 6/5
(d) 8/5
Solution:
(a) 2/5

Let A & T are events of selection of Aran and Taran respectively.
P(A) = 13
∴ P(A¯) = 1 – 13 = 23
P(T) = 15
∴ P(T¯) = 1 – 15 = 45
∴ Probability that only one will be selected = P(A). P(T¯) + P(A¯).P(T)
= 13 × 45 + 23 × 15 = 415 + 215 = 615 = 25

Question 36.
A company employed 7 CA’s, 6 MBA’s and 3 Engineer’s. Inhowmany ways the company can form a committee if the committee has two members of each type.
(a) 900
(b) 1,000
(c) 787
(d) 945
Solution:
(d) 945
The number of ways to make a committee containing two members of each type
= ⁷C₂ × ⁶C₂ × ³C₂
= 21 × 15 × 3
= 945

Question 37.
Two dice are thrown together. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other. 
(a) 2/3
(b) 1/6
(c) 1/3
(d) None of the above
Solution:
(b) 1/6
Sample space n(S) = 36
Let ‘E’ = ‘Events of getting a multiple of 2 on the 1st die and multiple of 3 on the IInd die
= {(2, 3) ; (2, 6) ; (4, 3) ; (4, 6) ; (6, 3) ; (6, 6)}
n(E) = 6
P(E) = n(E)n(S) = 636 = 16

Question 38.
The odds against A solving a certain problem are 4 to 3 and the odds in favour of B solving the same problem are 7 to 5 .What is the probability that the problem will be solved if they both try?
(a) 15/21
(b) 16/21
(c) 17/21
(d) 13/21 
Solution:
(b) 16/21

The odd against A solving a certain problem = 4 : 3
P(A) = Prob (to solve the problem) = 34+3 = 37
P(A¯) = prob (not to solve the problem) = 44+3 = 47
The odds in favour of B solving the same problem = 7 : 5
P(B) = Prob (to solve the problem) = 77+5 = 712
P(B¯) = Prob (not to solve the problem) = 57+5 = 512
Probability (the problem is solved)
1 – P(A¯∩B¯) = 1 – P(A¯¯¯¯)⋅P(B¯¯¯¯) = 1 – 47⋅512
Probability (problem is solved) = 1 – 521 = 1621

Question 39.
A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag 
(a) 10
(b) 12
(c) 14
(d) 16
Solution:
(b) 12

Let No. of Blue ball, = X
Total Ball in the Bag = (6 + X)
Prob of a Red ball P(R) = 6C16+xC1 = 66+x
and prob of a Blue Ball P(B) = xC16+xC1 = x6+x
Given, P(B) = 2P(R)
X(6+X) = 2×6(6+X) ⇒ X = 12
Tricks : GBC

Question 40.
The odds that a book will be received favourably by 3 independent reviewers are 5 to 2, 3 to 4, 4 to 3 respectively, then the probability that out of 3 critics the majority will be favorable is ______ 
(a) 209343
(b) 209434
(c) 209443
(d) 209350
Solution:
(a) 209343
(a) is correct

Question 41.
Find the probability of drawing spade on each of 2 consecutive draws from a well shuffled pack of cards when the draws are without replacement.
(a) 251
(b) 351
(c) 451
(d) 451
Solution:
(b) 351
(b) is correct
P(spades on consecutive draws of 2 cards)
= 1352c1×12c151c1 = 1352×1251
= 351 = 117

Question 42.
A bag contains 2 red 3 green and 2 blue balls. If 2 balls are drawn at random from the bag find the probability that none of them will be blue.
(a) 1121
(b) 57
(c) 1021
(d) 27 
Solution:
(c) 1021
(c) is correct
Sample Space = n ( S ) = ⁷C₂ = 21
p(None are blue) = 5c27c2 = 1021

Question 43.
If P(A) = 0.45, P(B) = 0.35, P(A and B) = 0.25 then P(A/B)
(a) 1.4
(b) 1.8
(c) 0.714
(d) 0.556 
Solution:
(c) 0.714
(c) is correct
P(A/B) = P(A∩B)P(B) = 0.250.35
= 0.714

Question 44.
Two coins are tossed simultaneously then the probability of getting exactly one head is 
(a) 34
(b) 23
(c) 14
(d) 12
Solution:
(d) 12
(d) is correct
P(Exactly 1 head) = P(H).P(T) + P(T).P(H)
= 12.12 + 12.12 = 24 = 12

Question 45.
The probability that a cricket team winning a match at Kanpur is 2/5 and loosing a match at Delhi is 1/7 .What is the probability of the team winning at least one match ? 
(a) 335
(b) 3235
(c) 1835
(d) 1735
Solution:
(b) 3235
P(At least 1 match win) = 1 – P (No match win)
= 1 – (1 – 25)17 = 1 – P (Loosing at Kanpur).P (Loosing at Delhi)
= 1 – 35.17 = 1 – 335 = 3235

Question 46.
For any two events
A₁, A₂; let P(A₁) = 23, P(A₂) = 38, P(A₁ ∩ A₂) = 14 then A₁, A₂ are
(a) Mutually Exclusive but not independent events
(b) Mutually Exclusive and iñdependent events
(c) Independent but not Mutually Exclusive
(d) None [1 Mark, June 2014]
Solution:
(c) Independent but not Mutually Exclusive
(c) is correct
(A₁ ∩ A₂) = 14 ≠ 0
So, A₁ and A₂ are not Mutually Exclusive Events.
P(A₁ ∩ A₂) = P(A₁).P(A₂)
= 23 × 38 = 14 (given)
∴ A₁ & A₂ are clearly Independent Events

Question 47.
If a pair of dice is thrown what is the probability of occurring neither 7 nor 11 ?
(a) 16
(b) 18
(c) 29
(d) 79 [1 Mark, June 2014]
Solution:
(d) 79
(d) is correct
n(s) = 6² = 36
Let A Event of getting Their sum = 7
= {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
n(A) = 6
Let B = Event of getting their sum = 11
= {(5, 6) ; (6, 5)}
n(B) = 2
P(A ∪ B) = n(A)+n(B)n(s)
[∵ A & B are mutually Exclusive Events]
= 6+236 = 836 = 29
P(Neither 7 nor 11) = P(A ∪ B)’
= 1 – p(A ∪ B)
= 1 – 29 = 79

Question 48.
An urn contains 2 red and 1 green balls, another urn contains 2 red and 2 green balls. An urn was selected at random and then a ball was drawn from it. ¡fit was found to be red then the probability that it has been drawn from first urn is
(a) 47
(b) 37
(c) 23
(d) 712 [1 Mark, June 2014]
Solution:
(a) 47
(a) is correct

Question 49.
If 6 coins are tossed simultaneously then the probability of obtaining exactly 2 heads is
(a) 164
(b) 6364
(c) 1564
(d) None 
Solution:
(c) 1564
(c) is correct
n = 6; p = 12(prob. of getting head)
q = 1- p = 1 – 12 = 12
p(x = 2) = ⁶C2.p².q⁴ = 15.(12)2⋅(12)4 = 1564

Question 50.
A die is thrown twice then the probability that the sum of the number is divisible by 4 is
(a) 19
(b) 13
(c) 1136
(d) 14 
Solution:
(d) 14 
(d) is correct
n(s) = 6² = 36
Let E = Event that the sum of numbers is divisible by 4.
= {(1, 3), (2, 2), (3, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}
n(E) = 9
∴ P(E) = n(E)n(S) = 936 = 14

Question 51.
There are 6 positive and 8 negative numbers. Four number are selected at random without replacement and multiplied. Find the probability that the product is positive. 
(a) 4201001
(b) 4091001
(c) 701001
(d) 5051001
Solution:
(d) 5051001
(d) is correct
Let 6 positive Nos. are 1,2, 3,4, 5, 6,
and 8 negative Nos. are -1,-2,-3, ……..; -8
∴ Sample space
= n(s) = ¹⁴C4 = =14!(4!)(10!)
= 14⋅13⋅12.11.10!4⋅3⋅2⋅1.10!
= 1001
Let Event = E = such that product of them is positive
n(E) = ⁶C₄ + ⁶C₂, ⁸C₂ + ⁸C₄ = (All + ve) (Two +ve & two -ve) + (All 4 are -ve)
= 15 + 15 × 28 + 70
= 505
∴ P(E) = n(E)n(S) = 5051001

Question 52.
P(A₁) = 3 /8; P(A₂) = 2 /3; P(A₁ ∩ A₂) = 1/4 then A₁ and A₂ will be
(a) Mutually exclusive & independent
(b) Exclusive but not independent
(c) Independent but not exclusive
(d) None 
Solution:
(c) Independent but not exclusive
(c) is correct
∵ P(A₁ ∩ A₂) = 14 (given) ≠ 0
∴ A₁ & A₂ are not Mutually Exclusive Events
P(A₂ ∩ A₂) = P(A₁).P(A₂) = 38×23 = 14
Clearly A₁ and A₂ are Independent Events

Question 53.
The sum of two numbers obtained in a single throw of two dice is ‘ S ’ .Then the probability of ‘s’ will be maximum when ‘S’ = 
(a) 5
(b) 7
(c) 6
(d) 8
Solution:
(b) 7
(b) is correct
S = Sum of face values of two due.
S = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
n(s) = 6 it is maximum if sum of their face is 7.

Question 54.
When an unbiased dice is rolled, find the odds in favour of getting of multiple of 3.
(a) 1/6
(b) 1/4
(c) 1/2
(d) 1/3 
Solution:
(c) 1/2
(c) is correct S = {1, 2, 3, 4, 5, 6}
Let E = {3, 6}; E’ = {1, 2, 3, 4, 5, 6} – {3, 6}
= {1, 2, 4, 5} ⇒ n(E) = 2; n(E’) = 4
Odds in favour of Event E = n(E)n(E1) = 24 = 12

Question 55.
Three coins are rolled, what is the probability of getting exactly two heads:
(a) 1/8
(b) 3/8
(c) 7/8
(d) 5/8 
Solution:
(b) 3/8
(b) is correct.
Given n = 3
P = Prob. (head) in 1 trial = 12
q = 1 – p = 1 – 12 = 12
P(X = 2) = ³C₂.p².q¹ = 3.(12)2⋅12 = 38

Question 56.
If a random sample of 500 Oranges produces 25 rotten oranges. Then the estimate of the proportion of rotten oranges in the sample is :
(a) 0.01
(b) 0.05
(c) 0.028
(d) 0.0593
Solution:
(b) 0.05
(b) is correct.
P = 25500 = 0.05

Question 57.
Two letter are drawn at random from word “ HOME” find the probability that there is no vowel.
(a) 5/6
(b) 1/6
(c) 1/3
(d) None 
Solution:
(b) 1/6
(b) is correct.
n(S) = ⁴C₂ = 6.
E = Event of no vowel = {H; M}
∴ n(E) = ²C₂ =1.
∴ P(E) = n(E)n(S) = 16

Question 58.
A bag contains 15 one rupee coins, 25 two rupee coins and 10 five rupee coins. If a coin is selected at random from the bag, then the probability of not selecting a one rupee coin is:
(a) 0.30
(b) 0.70
(c) 0.25
(d) 0.20 [1 Mark, Dec. 2015]
Solution:
(b) 0.70
(b) is correct.
P = 25+1050 = 0.70

Question 59.
If P(A) = 23, P(B) = 35, P(A ∪ B) = 56, thenP(A/B’)is 
(a) 712
(b) 512
(c) 14 (d)
(d) 12
Solution:
(b) 512

Question 60.
Two dice are tossed what is the probability that the total is divisible by 3 or 4
(a) 2036
(b) 2136
(c) 1436
(d) None
Solution:
(a) 2036
n(s)= 36
A = Event of getting Nos. such that the sum of their face Nos. is divisible by 3
= (sum 3 + sum 6 + sum 9 + sum 12)
= {(1, 2), (2, 1); (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
n(A) = 12
B = Event of getting Nos. such that the sum of their face Nos. is divisible by 4 = (sum 4 + sum 8 + sum 12)
= {(1, 3), (2,2); (3, 1); (2, 6), (3, 5), (4, 4); (5, 3); (6, 2); (6, 6)}
n(B) = 9
A ∩ B = Divisible by LCM of 3 & 4 = 12 = sum 12 = {(6, 6)}
n (A ∩ B) = 1
P(A ∪ B) = n( A)+n( B)−n( A∩B)n( S)
= 12+9−136 = 2036 = 59
(A) is correct.

Question 61.
If 2 dice are rolled simultaneously then the probability that their sum is neither 3 nor 6 is 
(a) 0.5
(b) 0.75
(c) 0.25
(d) 0.80
Solution:
(d) 0.80

Question 62.
In a game, cards are thorughly shuffled and distributed equally among four players. What is the probability that a specific player gets all the four kings ? 
(a) 52c4×48c1352c11
(b) 4c4×48c952c13
(c) 13c9×39c952c13
(d) 4c4×39c952c13
Solution:
(b) 4c4×48c952c13
Each candidate will get 13 cards.
∴ n(S) = ⁵²C₁₃
let E = Event of getting all 4 kings and rest 9 cards by a player out of 48.
∴ n(E) = ⁴C₄.⁴⁸C₉
∴ P(E) = n(E)n( S) = 4C448C952C13
(b) is correct

Question 63.
A bag contains 4 red and 5 black balls. Another bag contains 5 red, 3 black balls. If one ball is drawn at random from each bag. Then the probability that one red and one black ball drawn is _____.
(a) 1272
(b) 2572
(c) 3772
(d) 1372 
Solution:
Prob. (1 red and 1 black balls)
= P(R₁).P(B₂) + P(B₁).P(R₂)
Where R₁ = event of getting red ball from 1^st bag.
R₂ = event of getting red ball from 2^nd bag.
B₁ = event of getting black ball from 1^st bag.
B₂ = event of getting black ball from 2^nd bag.
∴ Prob. = 4C19C1×3C18C1+5C19C1×5C18C1
= 49×38+59×58 = 12+2572 = 3772

Question 64.
In a discrete random variable follows uniform distribution and assumes only the value 8, 9, 11 , 15 , 18, 20. Then P(x ≤ 15) is
(a) 1/2
(b) 1/3
(c) 2/3
(d) 2/7
Solution:
(c) 2/3
E = Event of Nos. ≤ 15 = {8, 9, 11, 15}
P(X ≤ 15) = n(E)n( S) = 46 = 23

Question 65.
A bag contains 6 green and 5 red balls. One ball is drawn at random. The probability of getting a red ball is?
(a) 511
(b) 611
(c) 56
(d) None 
Solution:
(a) 511
(a) is correct
n(S) = ¹¹C₁ = 11
Let E = Event of getting a red ball.
n (E) = ⁵C₁ = 5
P(E) = 511

Question 66.
If two events A, B P(A)= 12; P(B) = 13 and P(A∪B) = 23, then find P(A ∩ B)?
(a) 14
(b) 16
(c) 23
(d) 12 
Solution:
(b) 16
(b) is correct.
P(A ∩ B) = P(A) + P(B) – P(A∪B)
= 12 + 13 – 23 = 16

Question 67.
If P(A) = 23, P(B) = 38, (A ∩ B) = 14, then the even A & B are
(a) Independent and mutually exclusive
(b) Independent but not mutually exclusive
(c) Mutually exclusive but not independent
(d) Neither Independent nor exclusive 
Solution:
(b) Independent but not mutually exclusive
(b) is correct.
Since, P(A ∩ B) ≠ 0, S0 A & B are not mutually exclusive events.
Now, P(A ∩ B) = P(A). P(B) = 23.38 = 14
Hence , Events A & B are Independent events.
So, (b) is correct.

Question 68.
The probability of getting atleast one 6 from 3 throws of a perfect die is 
(a) 56
(b) (56)3
(c) 1 – (16)3
(d) 1 – (56)3 
Solution:
(d) 1 – (56)3 
(d) is correct
n = 3 (Trials)
P = Prob. of getting 6 in 1 trial = 16
q = 1 – p = 1 – 16 = 56
P(x ≥ 1) = 1 – P(x < 1)
= 1 – P (x = 0) = 1 – ³C₀. P⁰. q³
= 1 – 1.1.(5/6)³ = 1 – (5/6)³

Question 69.
For any two events A and B
(a) P(A – B) = P(A) – P(B)
(b) P(A – B) = P(A) – P(A ∩ B)
(c) P(A – B) = P(B) – P(A ∩ B)
(d) P(B – A) = P(B) + P(A ∩ B)
Solution:
(b) P(A – B) = P(A) – P(A ∩ B)

Question 70.
If P(A) = 23, P(B) = 14, P(A ∩ B) = 112, then P(BA) = ____
(a) 18
(b) 78
(c) 13
(d) 25
Solution:
(a) 18
(a) is correct
p(B/A) = P(A∩B)P(A)
= 1/122/3 = 112×32 = 18

Question 71.
For the events A & B if P(A) = 12,P(B) = 13 and P(A ∩ B) = 14 then P(AB) = 
(a) 1/2
(b) 1/6
(c) 2/3
(d) 3/4
Solution:
(d) 3/4

Question 72.
If A & B are two mutually exclusive events such that P(A ∪ B) = 23, P(A) = 25, then P (B) :
(a) 4/15
(b) 4/9
(c) 5/9
(d) 7/15
Solution:
(a) 4/15
P(A ∪ B) = P(A) + P(B)
[∵ A & B are two mutually exclusive events]
23 = 25 + P(B)
or; P(B) = 23 – 25 = 10−615 = 415

Question 73.
If a brother and a sister are applied for 2 vacancies in the same post. The probability that brother will select is 1/7 and that of sister is 1/5, then the probability that
(i) Both will select
(ii) Only one will select,
(iii) None of them will select: 
(a) 135, 1035, 2435
(b) 2435, 735, 1435
(c) 335, 2435, 1135
(d) 2435, 635, 2035
Solution:
(a) 135, 1035, 2435

Let A and B are events of selection of brother and sister respectively.
∵ Both events are independent.

(i) P(Both selected) = P(A∩B) = P(A) . P(B)
= 17 • 15 = 135

(ii) P(Only one will be selected]
= {P(A) – P(A ∩ B)} + {P(B) – P(A ∩ B)}
= 17 – 135 + 15 – 135
= 15 + 17 – 235
= 7+5−235 = 1035

(iii) P(None of them will be selected)
= P(A’) • P(B’) = (1 – 17) • (1 – 15)
= 67 × 45 = 2435

Question 74.
If 4 letters are put randomly among the 4 envelopes then the probability that all are not put in correct envelopes :
(a) 1/24
(b) 1
(c) 23/24
(d) 9/24 
Solution:
(c) 23/24
n(s) = 4! = 24
Let E = Events of putting letter in right envelop.
∴ n(E) = 1.1.1.1 = 1
P(E) = n(E)n(S) = 124
P(E’) = 1 – 1/24 = 23/24

Question 75.
Two broad divisions of probability are :
(a) Subjective probability and objective probability
(b) Deductive probability and mathematical probability
(c) Statistical probability and mathematical probability
(d) None of these 
Solution:
(a) Subjective probability and objective probability

Two broad divisions of Probability are

1. Subjective Probability
2. Objective Probability

Question 76.
The term “chance” and probability are synonyms :
(a) True
(b) False
(c) Both
(d) None 
Solution:
(a) True

Question 77.
The theorem of Compound Probability states that for any two events A and B
(a) P(A ∩ B) = P(A) × P(B/A)
(b) P(A ∪ B) = P(A) × P(B/A)
(c) P(A ∩ B) = P(A) × P(B)
(d) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) [CA (F) May 2018]
Solution:
(a) P(A ∩ B) = P(A) × P(B/A)
The theorem of Compound Probability states that for two events A and B
P(A ∩ B) = P(A) × P(B/A)

Question 78.
Variance of random variable x is given by
(a) E(X – µ)²
(b) E[X – E(X)]²
(c) E(X² – µ)
(d) (a) or (b)
Solution:
(d) (a) or (b)
Variance of a random variable x is given by
V(X) = E(X – µ)²
or
V(x) = E[X – E(X)]²
Note : ∵ µ = E(X)

Question 79.
What is the probability of having at least one ‘six’ perfect die?
(a) 5/6
(b) (5/6)³
(c) 1 – (1/6)³
(d) 1 – (5/6)³
Solution:
(d) 1 – (5/6)³

For a die Probability of getting Six
P(A) = 16 = p
P(A¯¯¯¯) = 1 – 16 = 56 = q
Here n = 3
P(getting at least ‘ 1 ’ Six) = P(X ≥ 1)
= 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – ³C₀.(16)0⋅(16)3−0
= 1 – 1 × 1 × (56)3
= 1 – (56)3

Question 80.
Sum of all probabilities of mutually exclusive and exhaustive events is equal to
(a) 0
(b) 1/2
(c) 1/4
(d) 1
Solution:
(d) 1

If events are mutually exclusive and exhaustive events then Sum of all probabilities = 1.

Question 81.
If P(A) = 12, P(B) = 13, and P(A ∩ B) = 14 then P(A ∪ B) is equal to
(a) 1112
(b) 712
(c) 1012
(d) 16 
Solution:
(b) 712
P(A ∪ B) = P(A) + P(B) – P(A∩B)
= 16 + 13 – 14 = 6+4−312 = 712

Question 82.
Two different dice are thrown simultaneously, then the probability, that the sum of two numbers appearing on the top of dice is 9 is 
(a) 19
(b) 89
(c) 79
(d) None of the above
Solution:
(a) 19
n(S) = 6² = 36
E = {(3, 6), (4, 5), (5, 4), (6, 3)}
n(E) = 4
∴ P(E) = n(E)n(S) = 436 = 19

Question 83.
If (A ∪ B) = 0.8 and P(A ∩ B) = 0.3 then P(A¯¯¯¯) + P(B¯¯¯¯) is equal to: 
(a) 0.3
(b) 0.5
(c) 0.9
(d) 0.7
Solution:
(c) 0.9

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
or; 0.8 = P(A) + (B) – 0.3
or; P(A) + P(B) = 0.8 + 0.3 = 1.1
∴ P(A¯¯¯¯) + P(B¯¯¯¯) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 1.1 = 0.9

Question 84.
The probability that a leap year has 53 Wednesday is
(a) 27
(b) 35
(c) 17
(d) 23 [1 Mark, Nov. 2018]
Solution:
(a) 27

1 Leap year = 366 days = 52 weeks & 2 days
S = {(Sun ; Mon); (Mon ; Tues);
(Tues ; Wed); (Wed ; Thurs) ;
(Thurs ; Fri); (Fri; Sat); (Sat; Sun)
∴ n(S) = 7
E = Event of getting Wednesday = {Tues ; Wed); (Wed ; Thurs)}
n(E) = 2
∴ P(E) = n(E)n(S) = 27

Question 85.
A coin is tossed six times, then the probability of obtaining heads and tails alternatively is
(a) 12
(b) 132
(a) 164
(a) 116
Solution:
(b) 132

P(Head & Tail alternatively) = P(H). P(T). P(H). P(T). P(H). P(T). + P(T) . P(H). P(T). P(H). P(T). P(H) = (12)6+(12)6
= 164 + 164 = 264 = 132

Question 86.
Ram is known to hit a target in 2 out of 3 shots where as Shyam is known to hit the same target in 5 out of 11 shots. What is the probability that the target would be hit if they both try?
(a) 911
(b) 611
(c) 1033
(d) 311
Solution:
(a) 911
P(Ram) = P(R) = 23
P(R’) = 1 – 23 = 13
and P(Shyam) = P(S) = 511
⇒ P(S’) = 1 – 511 = 611
∴ P (Target hit) = 1 – P (Target not hit)
⇒ 1 – P(R’).p(S’) = 1 – 13.611
= 1 – 211 = 911

Question 87.
The probability that a student is not a swimmer is 15, then the probability that out of five students four are swimmer is
(a) (45)4(15)
(b) 5C1(15)4(45)
(c) 5C4(45)4(15)
(d) None
Solution:
(c) 5C4(45)4(15)

Given that
let q = Prob. that a student is not a swimmer = 15
∴ P = Prob. (Swimmer) = 1 – q = 1 – 15 = 45
n = 5
P(X = 4) = ⁵C₄ • p⁴ • q¹
= 5C4⋅(45)4⋅(15)1

Question 88.
If Y ≥ x then mathematical expectation is
(a) E (X) ≥ E (Y)
(b) E (X) ≤ E (Y)
(c) E (X) = E (Y)
(d) E (X) . E (Y) = 1 
Solution:
(b) If y ≥ x then E (y) ≥ E (x)
E (x) ≤ E (y)

Question 89.
Two event A and B are such that they do not occurs simultaneously then they are called ____ events
(a) Mutually exhaustive
(b) Mutually exclusive
(c) Mutually independent
(d) Equally likely 
Solution:
(b) Mutually exclusive

Question 90.
According to Baye’s Theorem. P(E_k/A) = P(EK)P(A/EK)∑ni=1P(E1)P(A/E1)
(a) E₁, E₂ ………… are mutually exclusive
(b) P(E/A₁), P(E/A₂), ……… are equal to 1
(c) P(A₁/E), P(A₂/ E), …….. are equal to 1
(d) A & E₁’s are disjoint sets.
Solution:
(b) According to Baye’s Theorem

Question 91.

If a coin is tossed 5 times then the probability of getting Tail and Head Occurs alternatively is
(a) 18
(b) 116
(c) 132
(d) 164 
Solution:
(b) 116

P(getting tail and Head occurs Alternatives) = P(HTHTH) or P(THTHT)
= (12 × 12 × 12 × 12 × 12) + (12 × 12 × 12 × 12 × 12)
= 132 + 132 = (1+132)
= 232 = 116

Question 92.
When 2 – dice are thrown Simultaneously then the probability of getting at least one 5 is
(a) 1136
(b) 536
(c) 815
(d) 17 
Solution:
(a) If two dice are thrown then sample space n(S) = 36
Events ‘A’ = getting at least one ‘5’
‘A’ = {(5, 1) ; (5, 2); (5, 3) ; (5, 4); (5, 5) ; (5, 6) (1, 5) ; (2, 5); (3, 5); (4. 5) ; (6, 5)}
n (A) = 11
n(A) = n(A)n(S) = 1136