A and B are playing a game of chess. A can win or lose a round with probabilities (1/2) and (1/3) respectively. (1/6) is the probability to draw a round. The person who wins two round is the winner. What is the probability that B wins first round and Adoes not wins any round in first five round and A wins the game in exactly ten rounds?

Question

A and B are playing a game of chess. A can win or lose a round with probabilities (1/2) and (1/3) respectively. (1/6) is the probability to draw a round. The person who wins two round is the winner. What is the probability that B wins first round and Adoes not wins any round in first five round and A wins the game in exactly ten rounds? (A) 2((1/6))9 (B) 2((1/6))8 (C) 7((1/6))9 (D) 7((1/6))8

Tardigrade Admin · Accepted Answer

Probability =(1/3)((1/6))7 ⋅ 4 C 1((1/2)) × (1/2)=(2/68)

$2 (\frac{1}{6})^{9}$

$2 (\frac{1}{6})^{8}$

$7 (\frac{1}{6})^{9}$

$7 (\frac{1}{6})^{8}$

Probability $= \frac{1}{3} (\frac{1}{6})^{7} \cdot^{4} C_{1} (\frac{1}{2}) \times \frac{1}{2} = \frac{2}{6 ^{8}}$

2(61​)9

2(61​)8

7(61​)9

7(61​)8

Probability =31​(61​)7⋅4C1​(21​)×21​=682​

$2 (\frac{1}{6})^{9}$

$2 (\frac{1}{6})^{8}$

$7 (\frac{1}{6})^{9}$

$7 (\frac{1}{6})^{8}$

Probability $= \frac{1}{3} (\frac{1}{6})^{7} \cdot^{4} C_{1} (\frac{1}{2}) \times \frac{1}{2} = \frac{2}{6 ^{8}}$