This set of Engineering Mathematics Assessment Questions and Answers focuses on “Set Theory of Probability – 2”.

1. If P(^{B}⁄_{A}) = p(b), then P( A and B) =

a) p(b)

b) p(a)

c) p(b)p(a)

d) p(a) + p(b)

View Answer

Explanation: P(B /A) = p(b) implies A and B are independent events

Therefore, P(A and B) = p(a)p(b).

2. Two unbiased coins are tossed. What is the probability of getting at most one head?

a)^{1}⁄_{2}

b)^{1}⁄_{3}

c)^{1}⁄_{6}

d)^{3}⁄_{4}

View Answer

Explanation: Total outcomes = (HH, HT, TH, TT)

Favorable outcomes = (TT, HT, TH)

At most one head refers to maximum one head,

Therefore, Probability =

^{3}⁄

_{4}.

3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then

P(~A /~B) =

a) 1 – P(A /B)

b) P(~A)/P(~B)

c) Not Defined

d) (1 – P(A or B) ) /P(~B)

View Answer

Explanation: From definition of conditional probability we have

4. If A and B are two events, then the probability of exactly one of them occurs is given by

a) P(A ∩ B) + P( A ∩ B)

b) P(A) + P(B) – 2P(A) P(B)

c) P(A) + P(B) – 2P(A) P(B)

d) P(A) + P(B) – P(A ∩ B)

View Answer

Explanation: The set corresponding to the required outcome is

(A ∩ B) ∪ ( A ∩ B)

Hence the required probability is

P(A ∩ B) + P( A ∩ B).

5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(~M) + P(~N) is

a) 0.4

b) 1.2

c) 0.8

d) Indeterminate

View Answer

Explanation: Given : P(M or N) = 0.6, P(M and N) = 0.2

P(M or N) + P(M and N) = P(M) + P(N)

2 – ( P(M or N) + P(M and N) ) = 2 – ( P(M) + P(N) )

= ( 1 – P(M) ) + ( 1 – P(N) )

2 – (0.6 + 0.2) = P(~M) + P(~N) = 1.2.

6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red

View Answer

Explanation: Number of blue balls = y

Number of Red balls = x

Total number of balls = x + y

Probability of Blue ball first = y / x + y

No. of balls remaining after removing one = x + y – 1

7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is

a) 0.29

b) 0.37

c) 0.61

d) 0.75

View Answer

Explanation: Given: p(b) = 0.33, Pc) = 0.42

P(B and C) = 0.12

P(~B and ~C) = ?

P(~B and ~C) = 1 – P(B or C)

= 1 – p(b) – Pc) + P(B and C)

= 1 – 0.22 – 0.42 + 0.12

= 0.37.

8. A coin is biased so that its chances of landing Head is ^{2}⁄_{3} . If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is

a) ^{4}⁄_{27}

b) ^{8}⁄_{27}

c) ^{4}⁄_{9}

d) ^{2}⁄_{9}

View Answer

Explanation: Required probability =

^{2}⁄

_{3}x

^{2}⁄

_{3}x

^{1}⁄

_{3}=

^{4}⁄

_{27}.

9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?

a) ^{2}⁄_{7}

b) ^{5}⁄_{7}

c) ^{4}⁄_{21}

d) ^{17}⁄_{21}

View Answer

Explanation:Let H be the event of husband getting selected

W be the event of wife getting selected

Then, the event of neither of them getting selected is = (H ∩ W)

P (H ∩ W) = P (H) x P (W)

= ( 1 – P (H) ) x ( 1 – P (W) )

= ( 1 –

^{2}⁄

_{3}) x (1 –

^{3}⁄

_{7})

=

^{4}⁄

_{21}.

10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then

P (A|B) =

a) 0.5

b) 0

c) 0.25

d) 1

View Answer

Explanation: We know that,

P (A│B) = P(A ∩ B) /P(B)

= P((A ∪B) /P(B) )

= ( 1 – P(A ∪ B) ) /P(B)

= (1 – 0.5) /0.5

= 1.

11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?

a) ^{3}⁄_{4}

b) ^{1}⁄_{4}

c) ^{1}⁄_{2}

d) ^{1}⁄_{6}

View Answer

Explanation:S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

All 3 same outcomes mean either all head or all tail

Total outcomes = 8

Favourable outcomes = {HHH, TTT} = 2

∴ required probability =

^{2}⁄

_{8}=

^{1}⁄

_{4}.

12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour?

View Answer

Explanation:Total no.of balls = 5R+3Y = 8

No.of ways in which 2 balls can be picked =

^{8}C

_{2}

Probability of picking both balls as red =

^{5}C

_{2}/

^{8}C

_{2}

Probability of picking both balls as yellow =

^{3}C

_{2}/

^{8}C

_{2}

∴ required probability .

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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