To find the probability that at least 2 people on the committee are small business people using the hypergeometric distribution, we need to calculate the probability of selecting 2, 3, 4, or 5 small business people and sum them up.
First, we calculate the total number of ways to choose a committee of 5 people from the total population of 19 (11 civil employees + 8 small business people):
${{19}\choose{5}} = 11628$
Next, we calculate the number of ways to choose 2 small business people and 3 civil employees:
${{8}\choose{2}} * {{11}\choose{3}} = 1540 * 165 = 254100$
Then, we calculate the number of ways to choose 3 small business people and 2 civil employees:
${{8}\choose{3}} * {{11}\choose{2}} = 56 * 55 = 3080$
Next, we calculate the number of ways to choose 4 small business people and 1 civil employee:
${{8}\choose{4}} * {{11}\choose{1}} = 70 * 11 = 770$
Finally, we calculate the number of ways to choose all 5 small business people:
${{8}\choose{5}} = 56$
Adding all these possibilities together:
$254100 + 3080 + 770 + 56 = 258006$
So, the probability of selecting a committee with at least 2 small business people is:
$P = \frac{258006}{11628} \approx 0.0222$ or 2.22%
A committee of 5 people is to be formed from 11 civil employees and 8 small business people. what is the probability that at least 2 people are small business people is ? using hypergeometric distribution
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