Question

Two players A and B are competing at a trivia quiz game involving a series of questions. On any individual question, the probabilities that A and B give the correct answer are $\alpha$ and $\beta$ respectively, for all questions, with outcomes for different questions being independent. The game finishes when a player wins by answering a question correctly.
Compute the probability that A wins if B answers the first question.

• A. $\frac{\alpha }{1-(1-\alpha )(1-\beta )}$
• B. $\frac{(1-\beta )}{1-\left(\alpha \right)\left(\beta \right)}$
• C. $\frac{(1-\beta )\alpha }{1-(1-\alpha )(1-\beta )}$
• D. none of these

Answer 1

The correct option is C $\frac{(1-\beta )\alpha }{1-(1-\alpha )(1-\beta )}$

Let event A - A answers the first question;

event F - game ends after the first question;

event W - A wins.

To find:

$P\left(W/A^{\prime }\right)$

Now, clearly

$P\left(F/A\right)=P$ [A answers first question correctly] = $\alpha$,

$P\left(F^{\prime }/A\right)=1-\alpha ,$ and

$P(W/A\cap F)=1$, but $P(W/A\cap F^{\prime })=P\left(W/A^{\prime }\right)$, so that

$P\left(W/A\right)=P(W/A\cap F)P\left(F/A\right)+P(W/A\cap F^{\prime })P\left(F^{\prime }/A\right)$

$P\left(W/A\right)=(1\times \alpha )+\left(P\right(W/A^{\prime })\times (1-\alpha \left)\right)=\alpha +P\left(W/A^{\prime }\right)(1-\alpha )....\left(i\right)$

We have,

$P\left(F/A^{\prime }\right)=P$ [B answers first question correctly] = $\beta$,

$P\left(F^{\prime }/A\right)=1-\beta$

but $P(W/A^{\prime }\cap F)=0$. Finally $P(W/A^{\prime }\cap F^{\prime })=P\left(W/A\right)$, so that

$P\left(W/A^{\prime }\right)=(0\times \beta )+\left(P\right(W/A)\times (1-\beta \left)\right)=P\left(W/A\right)(1-\beta ).......\left(ii\right)$

Solving (i) and (ii) simultaneously gives, for B answers the first question,

$P\left(W/A^{\prime }\right)=\frac{(1-\beta )\alpha }{1-(1-\alpha )(1-\beta )}$