Question

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

• A. $p^2(2–p)$
• B. $p(2–p)$
• C. $p+p^2+p^3$
• D. $p^2(1–p)$

Answer 1

The correct option is A

$p^2(2–p)$

Explanation for the correct option:

Find the probability of candidate selection:

Candidate will be selected in following cases

$$\begin{gathered} i)PPP \\ ii)PPF \\ iii)PFP \\ iv)FPP \end{gathered}$$

Now,

$\begin{array}{rcl}P\left(i\right)& =& p\cdot p\cdot p\\ & =& p^3\end{array}$

$\begin{array}{rcl}P\left(ii\right)& =& p\cdot p\cdot (1-p)\\ & =& p^2-p^3\end{array}$

$\begin{array}{rcl}P\left(iii\right)& =& p\cdot (1-p)\cdot \frac{p}{2}\\ & =& \frac{p^2}{2}-\frac{p^3}{2}\end{array}$

$\begin{array}{rcl}P\left(iv\right)& =& (1-p)\frac{p}{2}\cdot p\\ & =& \frac{p^2}{2}-\frac{p^3}{2}\end{array}$

$\therefore$The required Probability $=p^3+(p^2-p^3)+\left(\frac{p^2-p^3}{2}\right)+\left(\frac{p^2-p^3}{2}\right)$

$$\begin{gathered} =2p^2-p^3 \\ =p^2(2-p) \end{gathered}$$

Hence, option ‘A’ is correct.