Answer to Question #264510 in Statistics and Probability for Muhammad Ali

Question

A hockey team wins with a probability of 0.6 and loses with a probability of 0.3.The team plays three games over the weekend. Find the probability that the team:

a)     wins all three games.

b)     wins at least twice and doesn’t lose.

c)      wins one game, loses one, and ties one (in any order).

(Hint: next match result is un affected by previous one

Answer 1

From the given information:

P(win) = 0.60, P(lose) = 0.30, P(tie) = 1 - P(win) - P(lose) =1 - 0.60 - 0.30 = 0.10

a)

The probability that team win all the three games is:

P(win)*P(win)*P(win) = 0.60 * 0.60 * 0.60 = 0.216

b)

Here we need to find the probability that team win 2 or 3 games and rest games were tie.

Number of ways of winning 2 games:

Number of ways of choosing 2 games out of 3 is C(3,2) =3

The probability that team will win 2 games and 3rd will be a tie is:

C(3,2) * P(win)*P(win) * P(tie) = 3 * 0.6*0.6*0.1 = 0.108

The probability that team will win at least 2 games and doesn't lose is

C(3,2) * P(win)*P(win) * P(tie) + P(win)*P(win) * P(win) =0.108 +0.216 = 0.324

(c)

Since order is not important so required probability is

P(win)P(lose)P(tie) = 0.6 *0.3 *0.1 = 0.018