Answer :
Answer:
78.8% probability the number of wins exceeds the number of losses.
Step-by-step 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]n = 3, p = 0.7[/tex]
What is the probability the number of wins exceeds the number of losses
It is the team winning 2 games, or all three. So
[tex]P(X \geq 2) = P(X = 2) + P(X = 3)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.7)^{2}.(0.3)^{1} = 0.441[/tex]
[tex]P(X = 3) = C_{3,3}.(0.7)^{3}.(0.3)^{0} = 0.343[/tex]
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) = 0.441 + 0.343 = 0.788[/tex]
78.8% probability the number of wins exceeds the number of losses.