At a game show, there are 6 people (including you and your friend) in the front row.The host randomly chooses 3 people from the front row to be contestants.The order in which they are chosen does not matter.There are 6C3 = 20 total ways to choose the 3 contestants.What is the probability that you and your friend are both chosen?A.B.C.D.
Question
At a game show, there are 6 people (including you and your friend) in the front row.The host randomly chooses 3 people from the front row to be contestants.The order in which they are chosen does not matter.There are 6C3 = 20 total ways to choose the 3 contestants.What is the probability that you and your friend are both chosen?A.B.C.D.
Solution
To solve this problem, we need to calculate the total number of ways to choose 3 contestants and the number of ways to choose you and your friend.
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Total number of ways to choose 3 contestants from 6 people: This is a combination problem, and the formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. So, the total number of ways to choose 3 contestants from 6 people is C(6, 3) = 6! / [3!(6-3)!] = 20.
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Number of ways to choose you and your friend: If you and your friend are both chosen, there is only one spot left for the remaining 4 people. So, the number of ways to choose you, your friend, and one other person is C(4, 1) = 4! / [1!(4-1)!] = 4.
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Probability that you and your friend are both chosen: This is the number of favorable outcomes (you and your friend are chosen) divided by the total number of outcomes. So, the probability is 4/20 = 0.2 or 20%.
So, the answer is 20%.
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