Question

Problem 3) A, B, and C are evenly matched tennis players. Initially A and B play a set and the winner then plays C. This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. Find the probability that A is the overall winner.

Answer 1

Answer:

the probability that A is the overall winner is 5/14 or 0.35714

Step-by-step explanation:

Given the data in the question;

probability that A is the overall winner P[A is the overall winner]

⇒ $P_{A}$$P_{A}$ + $P_{A}$$P_{C}$$P_{B}$$P_{A}$$P_{A}$ + $P_{A}$$P_{C}$$P_{B}$$P_{A}$$P_{C}$$P_{B}$$P_{A}$ $P_{A}$  + ................

= $P_{B}$$P_{C}$$P_{A}$$P_{A}$ + $P_{B}$$P_{C}$$P_{A}$$P_{B}$$P_{C}$$P_{A}$$P_{A}$ + $P_{B}$$P_{C}$$P_{A}$$P_{B}$$P_{C}$$P_{B}$$P_{C}$$P_{A}$ + ........

P(A) = $\frac{1}{2}$ = $P_{B}$ = $P_{C}$             ∴{ they are evenly matched player}

so P(A is overall winner) will be;

= ($\frac{1}{2}$)² + ($\frac{1}{2}$)⁵ + ($\frac{1}{2}$)⁸ + ..........+ ($\frac{1}{2}$)⁴ + ($\frac{1}{2}$)⁷ + ($\frac{1}{2}$)¹⁰ + .............

= ($\frac{1}{2}$)² [1 + ($\frac{1}{2}$)³ + ($\frac{1}{2}$)⁶ + .........] + ($\frac{1}{2}$)⁴[1 + ($\frac{1}{2}$)³ + ($\frac{1}{2}$)⁶ + .........]

= [ ($\frac{1}{2}$)² + ($\frac{1}{2}$)⁴ ] [ 1 / 1 - ($\frac{1}{2}$)³ ]

= [ ($\frac{1}{4}$) + ($\frac{1}{16}$) ] [ $\frac{8}{7}$ ]

= $\frac{5}{16}$ × $\frac{8}{7}$

= 5/14 or 0.35714

Therefore, the probability that A is the overall winner is 5/14 or 0.35714