Question

) 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.

Answer 1

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

$P(A)=P(1)+P(2)\\\\P(A)=p+(1-p)^{2}\times P(A)\\\\\therefore P(A)=\frac{p}{1-(1-p)^{2}}=\frac{1}{2-p}$

Part b)

We have

$P(A)=\frac{1}{2-p}$

Thus for least value put p = 0

Thus

$P(A)>\frac{1}{2-0}\\\\\therefore P(A)>1/2$