Question

Sol. [Key Concept: Three successive tests with given pass probabilities: combine independently.]

Let the probability of passing a test be p.

If a person passes a test, the probability of passing the next test remains p.

If a person fails a test, the probability of passing the next test becomes p/2.

For selection, he must pass at least two out of three tests.

Case 1: Pass all three tests

P(PPP) = p × p × p = p³

Case 2: Pass first two, fail third

P(PPF) = p × p × (1 − p) = p²(1 − p)

Case 3: Pass first, fail second, pass third

P(PFP) = p × (1 − p) × p/2

= p²(1 − p)/2

Case 4: Fail first, pass second and third

P(FPP) = (1 − p) × p/2 × p

= p²(1 − p)/2

Therefore,

P(selection) = p³ + p²(1 − p) + p²(1 − p)/2 + p²(1 − p)/2

= p³ + 2p²(1 − p)

= p³ + 2p² − 2p³

= 2p² − p³

Hence,