Question

A candidate takes 4 tests in succession. The probability that he passes first test is p. And the probability that he passes any successive tests is p or $\frac { \mathrm { p } } { 2 }$ according as he passes or fails in the previous test. He will be qualified only when he passes atleast 3 tests. Then the chance that he will be qualified.

• A.
• B.
• C.
• D. (1-p³)

Answer 1

Answer: (C)

Answer 2

If p stands for pass and F stands for Fail.

Then required ways, are FPPP, PFPP, PPFP, PPPF, PPPP.

Required probability = (1 – p) $\frac { \mathrm { p } } { 2 }$ x p x p + p x (1 – p) x p + p. p(1 – p) $\frac { \mathrm { p } } { 2 }$ + p.p.p.(1 – p) + p⁴

= (1 – p) + p³ (1 – p) + p⁴.