Players $A$ and $B$ play against each other in a racquet match consisting of at most $3$ sets.

Question

Players $A$ and $B$ play against each other in a racquet match consisting of at most $3$ sets. Each set is won by either Player $A$ or $B$, and the match is won by the first person to win two sets. Player $A$ has a probability of $\frac{a-k}{8}$, where $a$ is a real positive constant, of winning at the ${{k}^{\mathrm{th}}}$ set where $k=1$, $2$, $3$. Show that the probability that player $A$ will win the match is $\frac{-{{a}^{3}}+18{{a}^{2}}-59a+50}{256}$.[3] Find all the values of $a$ for the two players to have an equal chance of winning. Hence explain why only one value is suitable in this context.[3] Given that $a=7$ and player $A$ wins the match, find the probability that player $A$ wins the second set.[2]

Accepted Answer

(a)(i) $\mathrm P(A\text{ wins})=\dfrac{-a^3+18a^2-59a+50}{256}$ (ii) $a=-1,6,13$, with only $a=6$ suitable (b) $\dfrac{35}{44}$

2023 HCI P2 Q9

2023 HCI P2 Q9

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