1.

Three players, A, B and C, toss a coin cyclically in that order (that A, B, C, A, B, C, A, B . . .) Till a head shows. Let be the probability that the coin shows a head. Let α, β and γ be, respectively, the probabilities that A, B and C gets the first head. Prove that β = (1 – p)α. Determine α, β and γ (in terms of p)

Answer»

Given that p is the prob. that coin shows a head then 1 – p will be the prob. that coin shows a tail.

Now α = P (A gets the 1st head in 1st try)

⇒ α = P(H) + P(T) P(T) P(T) P(H) + P (T) P (T) P (T) P (T) P (T) P (T) P(H)

= p + (1 – p)3p + (1 – p)6p+ . . . . . . . . . . . . . . . . . . . . . 

= p[1 + (1 – p)3 (1 – p)6 + . . . . . . . . . . . . . . . . . . . . . . . ]

= p/1 – (1 – p)3 

NOTE THIS STEP . . . (i)

Similarly β = P (B gets the 1st head in 1st try) + P (B gets the 1st head in 2nd try) + . . . . . . . . . . . . . . . . . . 

= P(T) P(H) + p(T) p(T) p(T) p(T) p(T) P(H) + . . . . . . . . . . . . . . . . . . .

= (1 – p) p + (1 – p)4 + . . . . . . . . . . . . . . . . . . . . 

= (1 – p) p/1 – (1 – p)3 . . . . . . . . . . . . . . . . . . . . . (ii) 

From (i) and (ii) we get β = (1 – p)α

Also (i) and (ii) give expression for α and β in terms of p.

Also α + β + γ = 1(exhaustive events and mutually exclusive events)

⇒ γ = 1 – α – β = 1 - α – (1 – p) α

= 1 – (2 – p)α

= 1 – (2 – p)p/1 – (1 - p)3 

= 1 – (1 – p)3 – (2p – p2)/1 – (1 – p)3

= 1 – 1 + p3 + 3p(1 – p) – 2p + p2/1 – (1 – p)3

= p3 – 2p2 + p/1 – (1 – p)3

= p(p2 – 2p + 1)/1 – (1 – p)3

= p (1 – p) 2/1 – (1 – p)3


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