At a game show, there are 8 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 8C3 = 56 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 8 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 8C3 = 56 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 first need to determine the total number of ways in which the host can select 3 people out of 8. 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 out of 8 is C(8, 3) = 8! / [3!(8-3)!] = 56.
Next, we need to find the number of ways in which both you and your friend can be chosen. This means that the host is actually choosing 1 person out of the remaining 6 (since you and your friend are already chosen). So, the number of ways to do this is C(6, 1) = 6! / [1!(6-1)!] = 6.
Finally, the probability that both you and your friend are chosen is the number of favorable outcomes (you and your friend being chosen) divided by the total number of outcomes. So, the probability is C(6, 1) / C(8, 3) = 6 / 56 = 0.1071.
So, the answer is approximately 0.1071 or 10.71%.
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