Q108 / 160 · TS EAMCET 2014 · Mathematics
A candidate takes three tests in succession and the probability of passing the first test is $p$. The probability of passing each succeeding test is $p$ or $\frac{p}{2}$ according as he passes or fails in the preceding one. The candidate is selected, if he passes atleast two tests. The probability that the candidate is selected, is
- Correct answerA$p^2(2-p)$
- B$p(2-p)$
- C$p+p^2+p^3$
- D$p^2(1-p)$
Solution
Required probability
$=$ Probability of passing two test
+ Probability of passing all three test
$=P$ (passing $\mathrm{I}$ and $\mathrm{II}$ tests and fail in third test)
$+P$ (passing Ist test, fail in II test and passing
in IIIrd test)
+ $P$ (fail in I test, passing in IInd and IIIrd tests)
$+P$ (passing in all three tests)
$$
\begin{aligned}
& =n p q+p q \times \frac{p}{2}+q \frac{p}{2} \times p+p p p \\
& =p^2(1-p)+p(1-p) \frac{p}{2}+(1-p) \frac{p^2}{2}+p^3 \\
& =p^2\left(1-p+\frac{1}{2}-\frac{p}{2}+\frac{1}{2}-\frac{p}{2}+p\right)=p^2(2-p)
\end{aligned}
$$
Hence, the correct answer is option: 'A'.