Answer to Question #177862 in Statistics and Probability for Jessica

Question

1.     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 areαandβrespectively, 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

a)    A answers the first question,

b)    B answers the first question.

Answer 1

sample space S to be all possible infinite sequences of answers

even A- A answers the first question

event F- game ends after the first question

event W- A wins.

We want $P(W|A)$ and $P(W|\overline A )$

Using the Theorem of Total Probability, and the partition given by

$P(W|A) = P(W|A \cap F)P(F|A) + P(W|A \cap \overline F )P(\overline F |A)$

Now, clearly

$P(F|A) = \alpha ,\,\,P\left( {\overline F |A} \right) = 1 - \alpha$ and $P(W|A \cap F) = 1$ , but $P(W|A \cap \overline F ) = P(W|\overline A )$ , so that $P(W|A) = 1 \cdot \alpha + P(W|\overline A )\left( {1 - \alpha } \right) = \alpha + P(W|\overline A )\left( {1 - \alpha } \right)$ .

Similarly,

$P(W|\overline A ) = P(W|\overline A \cap F)P(F|\overline A ) + P(W|\overline A \cap \overline F )P(\overline F |\overline A )$

We have

$P(F|\overline A ) = \beta ,\,\,P(\overline F |\overline A ) = 1 - \beta$ , but

$P(W|\overline A \cap F) = 0$ . Finally

$P(W|\overline A \cap \overline F ) = P(W|A)$ , so that

$P(W|\overline A ) = 0 \cdot \beta + P(W|A)\left( {1 - \beta } \right) = P(W|A)\left( {1 - \beta } \right)$

a) $P(W|A) = \alpha + P(W|\overline A )\left( {1 - \alpha } \right) = \alpha + P(W|A)\left( {1 - \beta } \right)\left( {1 - \alpha } \right) \Rightarrow P(W|A) - P(W|A)\left( {1 - \beta } \right)\left( {1 - \alpha } \right) = \alpha \Rightarrow P(W|A)(1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)) = \alpha \Rightarrow P(W|A) = \frac{\alpha }{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}}$

Answer: $P(W|A) = \frac{\alpha }{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}}$

b) $P(W|\overline A )\left( {1 - \alpha } \right) = P(W|A) - \alpha \Rightarrow P(W|\overline A )\left( {1 - \alpha } \right) = \frac{\alpha }{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}} - \alpha = \frac{{\alpha - \alpha \left( {1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)} \right)}}{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}} = \frac{{\alpha \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}}{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}} \Rightarrow P(W|\overline A ) = \frac{{\alpha \left( {1 - \beta } \right)}}{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}}$

Answer: $P(W|\overline A ) = \frac{{\alpha \left( {1 - \beta } \right)}}{{1 - \left( {1 - \beta } \right)\left( {1 - \alpha } \right)}}$