Free Probability MCQs with Answers

• A. P(AorB) =0
• B. P(AandB)=0
• C. P(AnorB)=0
• D. P(AeitherB)=0

Explanation: Mutually exclusive events cannot occur together, so their intersection is empty and P(A and B) equals zero. Their union need not be zero because either event may still occur.

• A. Dependent
• B. Independent
• C. Overlapped
• D. None of These

Explanation: Mutually exclusive events cannot occur together, whereas independent events can occur together without affecting each other, so mutually exclusive events are conventionally described as dependent. This assumes both events have positive probability.

• A. It must have a value between 0 and 1
• B. It can be reported as a decimal or a fraction
• C. A value near 0 means that the event is not likely to occur/happens
• D. It is the collection of several experiment

Explanation: A probability is a numerical measure between 0 and 1 and may be written as a decimal or fraction; a value near 0 indicates an unlikely event. A collection of experiments is not a probability, but may be called a sample space or experiment set depending on context.

• A. Independent
• B. Mutually exclusive
• C. Bayesian
• D. Empirical

Explanation: The special multiplication rule, P(A∩B) = P(A)P(B), applies when events are independent. Mutually exclusive events cannot occur together, which is a different relationship and generally gives an intersection probability of zero.

• A. 4/36
• B. 13/36
• C. 21/36
• D. 15/36

Explanation: There are 11 outcomes with at least one 4 and 6 outcomes whose sum is 7; 2 outcomes belong to both groups. Thus the union has 11 + 6 − 2 = 15 outcomes out of 36, giving 15/36.

• A. Independent
• B. Mutually Exclusive
• C. Bayesian
• D. Empirical

Explanation: If occurrence of one event rules out the occurrence of another, their intersection is empty, so they are mutually exclusive. Independent events may occur together, so independence does not fit this description.

• A. 5
• B. 5×5
• C. 1
• D. None of these

Explanation: Choosing all 5 objects from 5 gives only one possible selection, so \(\binom{5}{5}=5!/(5!0!)=1\).

• A. 1.2.3.4,.........
• B. 1.2.3......(n-2) (n-1) (n)
• C. n. (n-1) (n-2)..............3.2.1.0
• D. None of these

Explanation: The product of the first n natural numbers is 1 × 2 × 3 × ⋯ × n, which is written as n!. Option b expresses this product in ascending order.

• A. 1/3
• B. 2/3
• C. 1/2
• D. 0

Explanation: Let P(T)=p, so P(H)=2p. Since P(H)+P(T)=1, 3p=1 and therefore P(H)=2/3.

• A. 4/5
• B. 1/20
• C. 0
• D. 1/9

Explanation: Using P(A|B)=P(A∩B)/P(B), we get 1/4=(1/5)/P(B). Rearranging gives P(B)=4/5.