# A student appears for tests I, II and III. The students is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2 , then p(1+q)=

Question : A student appears for tests I, II and III. The students is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2 , then p(1+q)=

(A) 1/2

(B) 1

(C) 3/2

(D) 3/4

Solution :

If A, B, C represent events that the student is successful in tests I, II, III, respectively. Then the
probability the student is successful is P[(A n B n C’) U (A n B’ n C) U (A n B n C)]

= P(A n B n C) + P(A n B n C) + P(A n B n C)

= P(A) P(B) P(C’) + P(A) P(B’) P(C) + P(A) P(B) P(C)

[therefore, A, B, C are independent events] \begin{aligned}\Rightarrow pq\left( 1-\dfrac{1}{2}\right) +p\left( 1-q\right) \left( \dfrac{1}{2}\right) +pq\left( \dfrac{1}{2}\right) \ \Rightarrow py+\dfrac{p}{2}-\dfrac{pq}{2}\ \Rightarrow \dfrac{pq}{2}+\dfrac{p}{2}\ \Rightarrow \dfrac{p}{2}\left( 1+q\right) \end{aligned}

therefore, \begin{aligned}\dfrac{p}{2}\left( 1+q\right) =\dfrac{1}{2}\ \Rightarrow p\left( 1+q\right) =1\end{aligned}

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