Answer to Question #175916 in Statistics and Probability for kmy

Question

Over the past 30 games between team A and team B, team A has won 14 times, team B has won 11 times and the game ended in a draw 5 times. If these 2 teams played 9 games this season, what is the probability that team A would win 5 games, team B would win 3 games, and the remaining game would be a draw.

Answer 1

Solution:

Probability of winning team A, $P(A)=\frac{14}{30}=\frac7{15}$

Probability of winning team B, $P(B)=\frac{11}{30}$

Probability of a draw, $P(D)=\frac{5}{30}=\frac16$

Let E be the event when A wins 5 games, B wins 3 games, and 1 draw out of 9 games.

Then, $P(E)=^9C_5(\dfrac{7}{15})^5 (\dfrac{8}{15})^4+^9C_3(\dfrac{11}{30})^3(\dfrac{19}{30})^6+^9C_1(\dfrac{1}{6})^1(\dfrac{5}{6})^8$

$\approx0.2256+0.2672+0.3488 \\=0.8416$