Probability: One at a Time - Outcome A AND Outcome B = Multiply Probabilities
What is the probability that a 4-person committee chosen at random from a group
consisting of 6 men, 7 women, and 5 children contains exactly one woman?
Thus, the total probability of picking exactly one woman is 4 times the probability of each of the scenarios.
>$$\text{Total probability} = 4\times \frac{11}{18}\times \frac{7}{17} \times \frac{10}{16} \times \frac{9}{15} $$
To calculate this, first reduce 4 with 16 and 9 with 18:
>$$\text{Total probability} = \frac{11}{2}\times \frac{7}{17}\times \frac{10}{4}\times \frac{1}{15}$$.
Then reduce the 10 with the 2 and 15:
>$$\text{Total probability} = \frac{11}{1}\times \frac{7}{17}\times \frac{1}{4}\times \frac{1}{3}$$.
Finally, calculate the result:
>$$\text{Total probability} = \frac{77}{204}$$.
Incorrect.
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Incorrect.
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Incorrect.
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Incorrect.
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Correct.
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First calculate the probability that a woman is chosen first and no other women chosen:
There are 7 women out of 18 people. Thus, 11 are not women. Break the committee down into four events—choosing a woman (W), followed by three events of choosing non-women (N):
>$$\text{Probability of WNNN} = \frac{7}{18} \times \frac{11}{17} \times \frac{10}{16} \times \frac{9}{15}$$.
Don't calculate yet. Think about the other scenarios. What about the probability of choosing a woman only in the second pick?
>$$\text{Probability of NWNN} = \frac{11}{18} \times \frac{7}{17} \times \frac{10}{16} \times \frac{9}{15}$$
Indeed. That's exactly the same as the first scenario since the probability of a woman being picked anywhere is identical.
$$\frac{77}{832}$$
$$\frac{1}{7}$$
$$\frac{308}{1{,}411}$$
$$\frac{11}{416}$$
$$\frac{77}{204}$$
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