Question

A and B play a game in which A's probability of winning is 2/3 in a series of seven games. What is probability that, A will win
(i) 6 or more games
(ii) From 3 to 4 games

Answer 1

Answer on Question #55990 – Math – Statistics and Probability

A and B play a game in which A's probability of winning is $2/3$ in a series of seven games. What is probability that, A will win

(i) 6 or more games

(ii) From 3 to 4 games

Solution

A and B play a game in which A's probability of winning is $2/3$ in a series of seven games.

(i) Probability that A will win 6 or more games is

$$\mathrm{P}(\geq 6) = \mathrm{P}(6) + \mathrm{P}(7) = \binom{7}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^1 + \binom{7}{7} \left(\frac{2}{3}\right)^7 \left(\frac{1}{3}\right)^0 = 7 \left(\frac{2}{3}\right)^6 \frac{1}{3} + \binom{2}{3}^7 = 0.263.$$

(ii) Probability that A will win from 3 to 4 games is

$$\mathrm{P}(3 \text{ or } 4) = \mathrm{P}(3) + \mathrm{P}(4) = \binom{7}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^4 + \binom{7}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^3 = \frac{7!}{3!4!} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^4 + \frac{7!}{4!3!} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^3 = 0.384.$$

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