Question

A football team wins its weekly game with probability 0.7. Suppose the outcomes of games on 3 successive weekends are independent. What is the probability the number of wins exceeds the number of losses

Answer 1

78.8% probability the number of wins exceeds the number of losses.

Explanation:

For each game, there are only two possible outcomes. Either the team wins, or they lose. Suppose the outcomes of games on 3 successive weekends are independent. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

`n = 3, p = 0.7`

What is the probability the number of wins exceeds the number of losses

It is the team winning 2 games, or all three. So

`P(X \geq 2) = P(X = 2) + P(X = 3)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(3,2).(0.7)^(2).(0.3)^(1) = 0.441`

`P(X = 3) = C_(3,3).(0.7)^(3).(0.3)^(0) = 0.343`

`P(X \geq 2) = P(X = 2) + P(X = 3) = 0.441 + 0.343 = 0.788`

78.8% probability the number of wins exceeds the number of losses.