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(D) Let $A, B$ and $C$ be the events that the student passes tests $I, II$ and $III$ respectively. The student is successful if he passes in $(I 	ext

Vedclass · Accepted Answer

All of the above

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(D) Let $A, B$ and $C$ be the events that the student passes tests $I, II$ and $III$ respectively. The student is successful if he passes in $(I 	ext{ and } II)$ or $(I 	ext{ and } III)$.This is represented by the event $(A \cap B) \cup (A \cap C)$.Using the inclusion-exclusion principle,$P((A \cap B) \cup (A \cap C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$.Since the events are independent,$P(A \cap B) = P(A)P(B) = pq$,$P(A \cap C) = P(A)P(C) = p(\frac{1}{2})$,and $P(A \cap B \cap C) = P(A)P(B)P(C) = pq(\frac{1}{2})$.Thus,the probability of success is $pq + \frac{p}{2} - \frac{pq}{2} = \frac{pq}{2} + \frac{p}{2} = \frac{p}{2}(q + 1)$.Given the probability of success is $\frac{1}{2}$,we have $\frac{p}{2}(q + 1) = \frac{1}{2}$,which simplifies to $p(q + 1) = 1$.If $p=1$,then $1+q=1 \Rightarrow q=0$. If $p=\frac{2}{3}$,then $\frac{2}{3}(q+1)=1 \Rightarrow q+1=\frac{3}{2} \Rightarrow q=\frac{1}{2}$.Since there are multiple pairs $(p, q)$ satisfying $p(q+1)=1$,there are infinitely many values for $p$ and $q$. Thus,all options $A, B$ and $C$ are correct.

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

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