Probability Test

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Probability Test - 4

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Question 1
1 / -0.25

The conditional probability of an event E ’s complement E ’, given the occurrence of the event F

A
P (E ′|F) = 1

B
P (E ′|F) = 1 –P (E|F)

C
P (E ′|F) = –1 + P (E|F)

D
P (E ′|F) = P (E|F)

Solution

As the total probability of an event is always 1 . therefore , P (E ′|F) = 1 –P (E|F).
Question 2
1 / -0.25

A coin is tossed three times, E : at most two tails , F : at least one tail. Find P(E|F)

A
4/7

B
3/7

C
2/7

D
6/7

Solution
Question 3
1 / -0.25

Given that the events A and B are such that P(A) = 1/2 , P (A ∪B) = 3/5 and P(B) = p. Find p if they are mutually exclusive

A
7/10

B
3/10

C
2/10

D
1/10

Solution
Question 4
1 / -0.25

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

A
3/52

B
5/52

C
1/52

D
1/4

Solution

Let
E₁ , E₂ and E₃ and are events of selection of a scooter driver , car driver and truck driver respectively.
Question 5
1 / -0.25

Which of the following conditions do Bernoulli trials satisfy?

A
Each trial has finite number of outcomes and the probability of each outcome remains the same in each trial.

B
Each trial has exactly two outcomes: success or failure and the probability of success or failure can vary in each trial.

C
Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.

D
Each trial has exactly three outcomes: success or failure and the probability of success or failure remains the same in each trial.

Solution

Bernoulli trials are true only when Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.
Question 6
1 / -0.25

If E, F and G are events then P ((E ∪F)|G) =

A
P (E|G) + P (G|F) –P ((E ∩F)|G)

B
P (E|G) + P (F|G) –P ((E ∩F)|G)

C
P (E|G) + P (F|G) –P ((E ∩F)|F)

D
P (G|E) + P (F|G) –P ((E ∩F)|E)

Solution

If E, F and G are events then P ((E ∪F)|G) represents the conditional probability of the given event . therefore P ((E ∪F)|G) = P (E|G) + P (F|G) –P ((E ∩F)|G) .
Question 7
1 / -0.25

A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

A
5/9

B
1/3

C
4/9

D
2/3

Solution
Question 8
1 / -0.25

Given that the events A and B are such that P(A) = 1/2 P (A ∪B) = 3/5 and P(B) = p. Find p if they independent.

A
1/5

B
1/3

C
1/4

D
1/2

Solution

Since A and B are independent events. Therefore
Question 9
1 / -0.25

A random variable is a real valued function whose domain is the.

A
set of irrational numbers

B
set of real numbers

C
sample space of a random experiment

D
set of integers

Solution

A random variable is a real valued function whose domain is the sample space of a random experiment .
Question 10
1 / -0.25

A die is thrown 6 times. If ‘getting an odd number ’is a success, what is the probability of 5 successes?

A
1/32

B
7/32

C
5/32

D
3/32

Solution

p = probability of success = 3/6 = ½. q = probability of failure = 1 –p = 1 –½= ½. let x be the number of successes , then x has the binomial distribution with : n = 6 , p = ½, q = ½.
Question 11
1 / -0.25

If E and F are events then P (E ∩F) =

A
P (E ∩F) P (F|E), P (E) ≠0

B
P (E) P (E|F), P (E) ≠0

C
P (E ∪F) P (F|E), P (E) ≠0

D
P (E) P (F|E), P (E) ≠0

Solution

If E and F are events then P (E ∩F) = P (E) P (F|E), P (E) ≠0.
Question 12
1 / -0.25

A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

A
5/9

B
1/9

C
7/9

D
4/9

Solution

n(S)=36.
Let A = event of getting sum 8.
= {(2,6),(3,5),(4,4),(5,3),(6,2),}
And B = event of getting a number less than 4 on red die.
={(1,1),(2,1),(3,1),(4,1),(5,1),(5,1), (1,2),(2,2),(3,2),(4,2),(5,2),(6,2), (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)}.
Question 13
1 / -0.25

Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4. Find P(A ∩B)

A
0.14

B
0.12

C
0.10

D
0.15

Solution

Let A and B be independent events with P (A) = 0.3 and P(B) = 0.4
Question 14
1 / -0.25

Let X be a random variable assuming values x1, x2,....,xn with probabilities p1, p2, ...,pn, respectively such that . Mean of X denoted by μ is defined as

A

B

C

D

Solution

Let X be a random variable assuming values x1, x2,....,xn with probabilities p1, p2, ...,pn, respectively such that Mean of X denoted by μ is defined as:
Question 15
1 / -0.25

A die is thrown 6 times. If ‘getting an odd number ’is a success, what is the probability of at least 5 successes?

A
7/64

B
5/64

C
9/64

D
3/64

Solution

p = probability of success = 3/6 = ½.
q = probability of failure = 1 –p = 1 –½= ½
. let x be the number of successes , then x has the binomial distribution with :
n = 6 , p = ½, q = ½.