Question

A student appears for tests I, II and III. The student is successful if he passes in tests I, II or I, III. The probabilities of the student passing in tests, I, II and III are respectively \(p, q\) and \(\frac{1}{2}\). If the probability of the student to be successful is \(\frac{1}{2}\). Then
(A) \(p(1+q)=1\)
(B) \(q(1+p)=1\)
(C) \(\mathrm{pq}=1\)
(D) \(\frac{1}{p}+\frac{1}{q}=1\)

Answer 1

Ans: (A)
Hint : Let \(x_{1}\) - He passes in test-I
\(\mathrm{x}_{2}-\mathrm{He}\) passes in test-II
\(x_{3}-\mathrm{He}\) passes in best - III
\(\mathrm{x}-\mathrm{He}\) is successful
\(x \equiv\left(x_{1} \cap x_{2} \cap x_{3}^{\prime}\right) \cup\left(x_{1} \cap x_{2}^{\prime} \cap x_{3}\right) \cup\left(x_{1} \cap x_{2} \cap x_{3}\right)\)
\(\therefore p(x)=p\left(x_{1}\right) \cdot p\left(x_{2}\right) \cdot p\left(x_{3}^{\prime}\right)+p\left(x_{1}\right) \cdot p\left(x_{2}^{\prime}\right) \cdot p\left(x_{3}\right)+p\left(x_{1}\right) \cdot p\left(x_{2}\right) \cdot p\left(x_{3}\right)\)
\(\Rightarrow \frac{1}{2}=p q \cdot \frac{1}{2}+p(1-q) \frac{1}{2}+p q \frac{1}{2} \Rightarrow p+p q=1 \therefore p(1+q)=1\)

Answer 2

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Answer 3

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