Question

A and B play a game taking alternative chances,the probability for a success is p and failure is q.If A takes the first chance, probability for his successis

• A. $\frac{p}{(1-q^2)}$
• B. $\frac{q}{(1-p^{-\prime })}$
• C. $p(1-p^2)$
• D. $\frac{1}{1-q^2}$

Answer 1

Answer: (A)

$\frac{p}{(1-q^2)}$

Let success be denoted by S and failure be denoted by F.
Then, for A to be successful, the outcomes of the game should be
$S,FFS,FFFFS,....$
i.e. A will be successful iff all previous turns result in failure.
$\therefore$ P(A is successful) = $p+q.q.p+q.q.q.q.p+.....\infty$
Probability = $p+q^{2}p+q^{4}p+...\infty$
= $\frac{p}{1-q^2}\because q<1$