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 at least two tests. The probability that the cndidate 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 $P^2(2-P)$

Candidate will be selected in following cases

i) P P P

ii) P P F

iii) P F P

iv) F P P

$P\left(I\right)=P\cdot P\cdot P=P^3$

$P\left(II\right)=P\cdot P\cdot (1-P)=P^2-P^3$

$P\left(III\right)=P\cdot (1-P)\cdot \frac{P}{2}=\frac{P^2}{2}-\frac{P^3}{2}$

$P\left(IV\right)=(1-P)\frac{P}{2}\cdot P=\frac{P^2}{2}-\frac{P^3}{2}$

$\therefore$ Probability of sucess $=P^3+(P^2-P^3)+\left(\frac{P^2-P_3}{2}\right)+\left(\frac{P^2-P^3}{2}\right)$

$=2P^2-P^3$

$=P^2(2-P)$