The probability that a certain football team wins a match is 2/3.Given that the team plays 4 matches, find the probability that the team wins 1)more than half of the number of matches played

The probability is 32/81

Step-by-step explanation:

We know that:

The probability that the team wins is p = 2/3

Then the probability of not winning is q = 1 - p = 1 - 2/3 = 1/3

The team plays 4 matches.

We want to find the probability that the team wins 1 more than half of the number of matches played.

So if there are 4 matches, the half is 2 matches

one more than half is then 3 matches.

Let's assume that the team wins the first 3 matches, and does not win the last match.

match one, the team wins with prob: P₁ = 2/3

match two, the team wins with prob: P₂ = 2/3

match three, the team wins with prob: P₃ = 2/3

match four, the team wins with prob: P₄ = 1/3

The joint probability is the product of all the individual probabilities:

P = (2/3)*(2/3)*(2/3)*(1/3)

Now we must also considerate the permutations, we have the cases:

the team does not win the fourth match

the team does not win the third match

the team does not win the second match

the team does not win the first match

So we have just four permutations, then the total probability is:

probability = 4*P = 4*(2/3)*(2/3)*(2/3)*(1/3) = 4*(8/81) = (32/81)

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: