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) Two players, A and B, alternatively and independently toss a coin and the first player to obtain a head wins. Assume, player A tosses first and that P(‘head’) = p. (a) What is the probability that A wins? (b) Show that for all p ∈ (0, 1), P("A wins") > 1/2.

User Neerajkh
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1 Answer

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Answer with Step-by-step explanation:

Part a)

Player A wins as follows:

1) He wins during the first toss itself.

2)Player A losses and then Player B also losses his turn and Player A wins subsequently

Probability of event 1 is 'p' (Given)

thus we have


P(A)=P(1)+P(2)\\\\P(A)=p+(1-p)^(2)* P(A)\\\\\therefore P(A)=(p)/(1-(1-p)^(2))=(1)/(2-p)

Part b)

We have


P(A)=(1)/(2-p)

Thus for least value put p = 0

Thus


P(A)>(1)/(2-0)\\\\\therefore P(A)>1/2

User Sugandika
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