Statistics

Statistical Probability and Expected Value – MCQ

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Last Updated on: 17th June 2025, 11:38 am

Statistical Probability – MCQ

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1. Each outcome of a random experiment is called
(a) Primary event
(b) Probable event
(c) Complementary event
(d) None of them

Each outcome of a random experiment is called Elementary event or Sample point. So, Option (d) is correct

2. All possible outcomes of a random experiment forms the
(a) Events
(b) Sample event
(c) Simple space
(d) Compound event.

All ;possible outcome of a random experiment is called Sample Even. So, Option (d) is correct

Option (b) is correct

3. If two events cannot occur simultaneously in the same trial then they are
(a) Simple events
(b) Favourable events
(c) Mutually exclusive events
(d) Compound events.
Two or more events are said to be equally likely if events are not possible at a time.

Option (c) is correct

4. The probability of the intersection of two mutually exclusive events is always
(a) – 1
(b) 0
(c) 1
(d) 100
Mutually exclusive events can not occur simultaneously. So, there is no chance of intersection of two mutually exclusive events. So probability is zero.

Option (b) is correct

5. If P(A) = 0, then  the event A
(a) Will never happen
(b) Will always happen
(c) May happen
(d) May not happen
Probability A = 0 means that event A will never happen.

Option (a) is correct

6. When P(A) = 1, then event A is called.
(a) Symmetric event.
(b) Dependent event.
(c) Improbable event.
(d) Sure event
Probability A = 1 means. Event A will surely happen. So P(A) = 1 implies event A is a sure event.

Option (d) is correct

7. If A, B and C are mutually exclusive and exhaustive events, then (P(A) + P(B) + P(C) equals to
(a) \displaystyle \frac{1}{3}
(b) 1
(c) 0
(d) Any value between 0 and 1
If A, B and C are mutually exclusive and exhaustive events then P(A) + P(B) + P(C) = 1

Option (b) is correct

8. Probability can take values from.
(a) 0 to 1
(b) – 1 to 1
(c) – 0 to 2
(d) – 1 to 2.

The value of probability lies between 0 to 1

Option (a) is correct

9. Probability is expressed as
(a) Percentage
(b) Ratio
(c) Proportion
(d) All the above

Probability may be expressed as Percentage, ratio or Proportion

Option (d) is correct

10. When two events A and B are independent, then P(A \displaystyle \cap B)

(a)    equals to P(A) + P(B)
(b)    equals to P(A) x P(B)
(c)    equals to P(A) x P(B/A)
(d)    equals to P(B) + P(A/B).

Options Analysis:

(a) Equals to P(A)+P(B)
Explanation: This describes the sum of probabilities of two events, which applies to mutually exclusive events. It is not applicable for independent events where probabilities can overlap in their outcomes.
So, this option is incorrect. Independence deals with the multiplication of probabilities, not addition.

(b) Equals to P(A)×P(B)
Explanation: This is the correct formula for the probability of the intersection of two independent events. When A and B are independent, the occurrence of A does not affect the occurrence of B, and their probabilities multiply.
So, this option is Correct. It follows the definition of independent events: P(A∩B)=P(A)×P(B).

(c) Equals to P(A)×P(B/A)
Explanation: While P(B/A)XP(B/A) represents the conditional probability of B given A, for independent events, P(B/A)=P(B)P(B/A) = P(B)P(B/A)=P(B). Hence, this simplifies to P(A)×P(B)P(A)
So, this option is Partially Correct. Though technically correct, this is not the most direct answer for independent events, as P(B/A)XP(B/A) is unnecessary to compute for P(A∩B)

(d) Equals to P(B)+P(A/B)
Explanation: This adds the probability of B and the conditional probability of A given B. For independent events, P(A/B)=P(A)P(A/B) = P(A)P(A/B)=P(A), but this option incorrectly uses addition instead of multiplication.
So, this option is Incorrect. Addition is not used for finding the intersection probability.
For independent events A and B, their probabilities satisfy the relationship: P(A∩B)=P(A)×P(B).
This reflects the definition of independence, as the occurrence of one does not influence the probability of the other.

So, option (b) is correct.

11. If A and B are two events, the probability of occurrence of either A or B is given as
(a) P( A) + P(B)
(b) P(A \displaystyle \cup B)
(c) P(A) P(B)
(d) P(A B)

In probability, the phrase “occurrence of either A or B” (or “A or B” for short) signifies the event where at least one of the events A or B happens. This is precisely the definition of the union of two events.

  • Union of Events (A ∪ B): This represents the event that A occurs, or B occurs, or both A and B occur.

Let’s look at other options:

(a) P(A) + P(B): This formula is used for the probability of A or B only if A and B are mutually exclusive events (meaning they cannot happen at the same time). In general, for any two events, you’d need to subtract the probability of their intersection to avoid double-counting: P(A ∪ B) = P(A) + P(B) – P(A ∩ B).

(c) P(A) P(B): This formula is used for the probability of A and B only if A and B are independent events. This represents the probability that both A and B occur.

(d) P(A B): This notation is ambiguous and not standard in probability theory.

The correct option is (b) P(A ∪ B).

12. If A and B are two events, the probability of occurrence of A and B simultaneously is given as
(a) P( A) + P(B)
(b) P(A \displaystyle \cup B)
(c) P(A B)
(d) P(A) P(B)

The phrase “occurrence of A and B simultaneously” refers to the event where both A and B happen at the same time. This is precisely the definition of the intersection of two events.
Intersection of Events (A ∩ B): This represents the event that both A occurs AND B occurs. In the context of the options provided,
P (A ∩ B): is often written as P(A B) or P(AB) as a shorthand for the intersection, though P(A ∩ B) is the more formal and universally recognized notation.
Assuming (C) P(A B) is intended to represent P(A ∩ B), it is the correct answer.
Let’s look at other options:
• (A) P(A) + P(B): This formula is used for the probability of A or B only if A and B are mutually exclusive events (meaning they cannot happen at the same time).
• (B) P(A U B): This represents the probability of the union of A and B, meaning the probability that A occurs, or B occurs, or both occur (“either A or B”).
• (D) P(A) P(B): This formula is used for the probability of A and B simultaneously only if A and B are independent events. While it calculates the probability of both occurring, it’s a specific case and not the general definition for “A and B simultaneously” for any two events. The general definition is the intersection, P(A ∩ B).
(c) P(A ∩ B) is Correct expression for simultaneous occurrence.

13. If P(A ∩ B) = 0, then the two events A and B are
(a) Mutually exclusive
(b) Exhaustive.
(c) Equal likely.
(d) Independent
P (A \displaystyle \cap B) = 0 denotes that events A and event B can not occur simultaneously. So, event A and event B are mutually exclusive event.

Option (a) is correct

14. For two events A and B, P(A\displaystyle \cap B) = P(A) + P(A) only if
(a) A and B are equally likely events
(b) A and B are exhaustive events
(c) A and B are mutually independent
(d) A and B are mutually exclusive.

The correct option is (c) A and B are mutually independent.

Reason:

Let’s analyze the condition P(A∩B)=P(A) in the context of different types of events:

Understanding P(A ∩ B) = P(A): This equation implies that the probability of both A and B occurring is the same as the probability of A occurring alone. This can only happen if the occurrence of B does not somehow “reduce” the probability of A’s occurrence in their joint happening, and it usually means that A is a subset of B, or that B is certain when A occurs, or specifically in the context of independence.

Let’s examine each option:

  • (a) A and B are equally likely events: This means P(A)=P(B). This condition alone does not tell us anything about their intersection. So, this is incorrect.
  • (b) A and B are exhaustive events: This means P(A∪B)=1. This implies that at least one of the events must occur. This condition doesn’t directly lead to P(A∩B)=P(A)..
  • (c) A and B are mutually independent: If A and B are mutually independent, then by definition, P(A∩B)=P(A)P(B).
  • (d) A and B are mutually exclusive: This means A∩B=∅, so P(A∩B)=0. If P(A∩B)=P(A), then it would imply P(A)=0. This is a very specific case and not generally true for any mutually exclusive events. So, this is incorrect.

Therefore, the condition P(A∩B)=P(A) is most directly and generally related to the concept of independence, particularly when one event (B) is a certain event, or if A is a subset of B and they are independent. The correct option is (c) A and B are mutually independent .

15. Addition Theorem of probability states that for any two events A and B
(a) P(A \displaystyle \cup B) = P(A) + P(B)
(b) P(A \displaystyle \cup B) = P(A) + P(B) + P(A \displaystyle \cap B)
(c) P(A \displaystyle \cup B) = P(A) + P(B) – P(A \displaystyle \cap B)
(d) P(A \displaystyle \cup B) = P(A) x P(B)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) is Correct

Option (c) is correct

16. If an unbiased coin is tossed twice, the probability of obtaining at least one tail is

(a) 0.25
(b) 0.50
(c) 0.75
(d) 1.00

a) 0.25 – P (both heads).
(b) 0.50 – Miscalculated.
(c) 0.75 – Correct: only HH is excluded. P( 3/4)
(d) 1.00 – Implies certainty. 

Option (b) is correct

17. In a single throw of two dice, find the probability of getting a total of 3 or 5
(a) \displaystyle \frac{1}{3}

(b) \displaystyle \frac{1}{4}

(c) \displaystyle \frac{1}{6}

(d) none of these
A = (1, 2), (2, 1), (2, 3), (3, 2), (1, 4), (4, 1)

\displaystyle \therefore P(A) = \displaystyle \frac{6}{{36}}=\frac{1}{6}

The question is Two dice: P(sum = 3 or 5)

P(1/6)  is Correct; as 6 possible  out of 36.  ..

Option (c) is correct

18. Two dice are thrown. The probability of getting an odd number on one and a multiple of three on the other is
(a) \displaystyle \frac{{11}}{{36}}

(b) \displaystyle \frac{9}{{36}}

(c) \displaystyle \frac{7}{{36}}

(d) none of these
Odd numbers in a dice are 1,3,5
Multiple of 3 in dice are 3,6
A = (1, 3), (3, 3), (5, 3), (1, 6), (3, 6), (5, 6), (3, 1), (3, 5), (6, 3), (6, 5), (6, 1)
\displaystyle \therefore P(A) = \displaystyle \frac{{11}}{{36}}
Odd numbers are 1, 3, 5,

So, Option (a) is correct

19. A and B are mutually exclusive events of an experiment. If P(not A) =0.65, P(A È B) = 0.65 and P(B) = P, Then the value of p is  
(a) 0.45
(b) 0.30
(c) 0.25
(d) None of these
P (not A) = .65
P(A) = 1 – .65 = .35
P(A \displaystyle \cup B) = P(A) + P(B)
.65 = .35 + P(B)
Or, P(B) = .65 – .35 = .30
P(B) = P = .30

So, Option (b) is correct

20. Probability mass function is always:
(a) 0
(b) Greater than 0
(c) Greater than or equal to 0
(d) Equally likely.

So, Option (c) is correct

21. Probability of the entire sample space is
(a) 1
(b) 1/2
(c) 0
(d) None of these.

Probability of entire sample space always equals 1.

So, Option (a) is correct

22. The sum of probability mass function is equal to:
(a) 1
(b) 0
(c) – 1
(d) None of these.

A Probability Mass Function (PMF) describes the probability of each possible outcome for a discrete random variable. Sum of all probabilities = 1…

So, Option (a) is correct

23. If events A and B are mutually exclusive, the probability that either A or B occurs is given by:
(a) P(A + B) = P(A) – P(B)
(b) P(A + B) (A) + P(B) – P(AB)
(c) P(A + B) = P(A) – P(B) + P(AB)
(d) P(A + B) = P(A) + P(B)

(a) P(A + B) = P(A) × P(B) –  Incorrect; product rule applies only to independent events, not mutually exclusive.

(b) P(A + B) = P(A) + P(B) – P(A ∩ B) –  This is for non-mutually exclusive events.

(c) P(A + B) = P(A) × P(B) + P(A ∩ B) –adds a product term unnecessarily. So Invalid option

(d) P(A + B) = P(A) + P(B) – for mutually exclusive events, the intersection is 0. So, P (A Ç B) = 0. So, P (A + B) = P (A) + P (B).   

So, Option (d) is correct

24. The probability that A can solve a problem is 2/3 and that B can solve is 3/4. If both of them attempt the problem, what is the probability that the problem get solved?
(a) 9/12
(b) 7/12
(c) 5/12
(d) 11/12
Probability of A will solve the problem P (A) = \displaystyle \frac{2}{3}

Probability of B will solve the problem P (B) = \displaystyle \frac{3}{4}

Probability of both A and B will solve the problem
P (A \displaystyle \cap B) = P (A). P (B) = \displaystyle \frac{2}{3}\times \frac{3}{4}=\frac{1}{2}

Probability of either A or B will solve the problem
P (A \displaystyle \cap B) = P (A) + P (B) – P (A \displaystyle \cap B)

\displaystyle \frac{2}{3}+\frac{3}{4}-\frac{1}{2}

\displaystyle =\frac{{11}}{{12}}

So, Option (d) is correct

25. A and B are events and P(A) = 0.4, P(A \displaystyle \cup B) = 0.7. If A and B are independent, then P (b) is:
(a) 0.22
(b) 0.33
(c) 0.3
(d) 0.5
P (A) = .40 P(A \displaystyle \cap B) = .70
A & B are independent event
\displaystyle \therefore P(A \displaystyle \cap B) = P (A) + P (B) – P(A \displaystyle \cap B)
Or, .70 = .40 + P (B) – P (A). P (B)
(Incase of independent event P (A \displaystyle \cap B) = P (A). P (B)
Or, .70 – .40 = P (B) – P (A). P (B)
Or, .30 = P (B) {1 – P (A)}
Or, .30 = P (B) (1 – .40)
Or, .30 = P (B) x .60
Or, P (B) = \displaystyle \frac{{.30}}{{.60}}=.50

So, Option (d) is correct

26. The probability of choosing at random, a number that is divisible by 6 or 8 from among 1 to 90 is:
(a) 1/6
(b) 11/90
(c) 1/30
(d) 23/90
Probable event = [ 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 8, 16, 32, 40, 53, 64, 80, 88]
Favourable events = 23
Total events = 90
\displaystyle \therefore Required probability = \displaystyle \frac{{23}}{{90}}

So, Option (d) is correct

27. The probability of two events A and B are 0.25 and 0.40 respectively. The probability that both A and B occur is 0.15. The Probability neither A nor B occur, is:
(a) 0.35
(b) 0.65
(c) 0.5
(d) 0.75
P (A) = .25 P (B) = .40
P (A \displaystyle \cup B) = P (A) + P (B) – P (A \displaystyle \cap B)
= .25 + .40 – .15
= .50
The Probability neither A nor B occur, is
P (A’ \displaystyle \cap B’) = 1 – .50 = .5

So, Option (c) is correct

28.The probability of getting qualified in IIT JEE and AIEEE by a student are respectively \displaystyle \frac{1}{5} and \displaystyle \frac{3}{5}. The probability that the student gets qualified for one of these tests is:

(a) \displaystyle \frac{3}{{25}}

(b) \displaystyle \frac{{17}}{{25}}

(c) \displaystyle \frac{{22}}{{25}}

(d) \displaystyle \frac{8}{{25}}
Event A = IIT JEE
Event B = AI EEE
P (A) = \displaystyle \frac{1}{5} P (B) = \displaystyle \frac{3}{5}

Probability of selection in both test
= P (A \displaystyle \cap B) = P (A). P (B)

\displaystyle \frac{1}{6}\times \frac{1}{3}=\frac{3}{{25}}

Probability of selection in any one test
P (A \displaystyle \cup B) = P (A) + P (B) – P (A \displaystyle \cap B)

\displaystyle \frac{1}{5}+\frac{3}{5}-\frac{3}{{25}}

=\displaystyle \frac{4}{5}-\frac{3}{{25}}

\displaystyle =\frac{{20-3}}{{25}}=\frac{{17}}{{25}}

So, Option (b) is correct

29. If E is an event, that   P(E) is equal to
(a) P(E)
(b) 1 – P(E)
(c) – P(E)
(d) 1 + P(E).

As  per rule of probability, Complement of a probability (means not occurring the probable event) = 1 – P(E)  

So, Option (b) is correct

30. Given two mutually exclusive event A and B, such that P(A) = 0.45 and P(B) = 0.35, then P(A or B) =
(a) 0.5
(b) 0.05
(c) 0.25
(d) 0.8
Event A and event B are mutually exclusive
\displaystyle \therefore P (A or B) = P (A) + P (B)
= .45 + .35
= .80

So, Option (d) is correct