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A doctor has 2 doses of flu protection vaccine left. He has 7 women and 10 men who want the medication. If the names of 2 of these people are selected at​ random, determine the probability that 2 ​men's names are selected. The problem is to be done without replacement. Use combinations to determine the probability.

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Final answer:

The probability that two men's names are chosen without replacement from a group of 7 women and 10 men is approximately 33.09%.

Step-by-step explanation:

The question asks us to find the probability that two men's names are selected without replacement from a group of 7 women and 10 men when two doses of a flu protection vaccine are available. To solve this problem, we use combinations to determine the probability.

First, we find the total number of ways to choose 2 people from the group of 17 (7 women + 10 men), which is a combination of 17 taken 2 at a time, denoted as C(17, 2). Next, we find the number of ways to choose 2 men from the 10 available, which is C(10, 2). The probability of selecting 2 men is then calculated by dividing the number of ways to choose 2 men by the total number of ways to choose 2 people.

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. We use this to calculate:

  • C(17, 2) = 17! / (2!(17 - 2)!) = 136
  • C(10, 2) = 10! / (2!(10 - 2)!) = 45

The probability that the two names selected are men's names is C(10, 2) / C(17, 2) = 45 / 136 = 0.33088, or about 33.09%.

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