PUNE MCA Exam 2015 Question Paper with answers for online practice

Concept:Pairwise independence requires each pair of events to satisfy $P(A \cap B) = P(A)P(B)$, but full independence also requires $P(A \cap B \cap C) = P(A)P(B)P(C)$.Explanation:• Total bulbs = 100 (50 defective, 50 non-defective).• With replacement, each draw has $P(\text{defective}) = 0.5$, $P(\text{non-defective}) = 0.5$.• Event $X$: first bulb defective, so $P(X) = 0.5$.• Event $Y$: second bulb non-defective, so $P(Y) = 0.5$.• Event $Z$: both defective or both non-defective. $P(Z) = P(\text{both defective}) + P(\text{both non-defective}) = 0.25 + 0.25 = 0.5$.• Check $X$ and $Y$: $P(X \cap Y) = 0.5 \times 0.5 = 0.25 = P(X)P(Y)$ → independent.• Check $X$ and $Z$: $X \cap Z$ means first defective and both defective → $P(X \cap Z) = 0.25 = P(X)P(Z)$ → independent.• Check $Y$ and $Z$: $Y \cap Z$ means second non-defective and both non-defective → $P(Y \cap Z) = 0.25 = P(Y)P(Z)$ → independent.• Thus $X, Y, Z$ are pairwise independent.• Check full independence: $X \cap Y \cap Z$ is impossible (first defective and second non-defective cannot satisfy both defective or both non-defective), so $P(X \cap Y \cap Z) = 0$, while $P(X)P(Y)P(Z) = 0.125$ → not equal, so not fully independent.• They are not mutually exclusive because $X$ and $Y$ can both occur ($P > 0$).Answer:The events are pairwise independent.