Probability: 'At Least' Questions
What is the probability that a four-person committee chosen at random from a group
consisting of 6 men, 7 women, and 5 children contains at least one woman?
Correct.
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"At least one woman" allows 1, 2, 3, or 4 women on the committee. Therefore, the forbidden scenarios in this question are only those in which _no woman is chosen_ among the four people. Calculate the probability of that happening. There are 7 women in 18 people, which means that 11 are not women:
>$$\text{Forbidden probability} = \frac{11}{18}\times \frac{10}{17}\times \frac{9}{16}\times \frac{8}{15}$$
>>>>>$$\ \ \ = \frac{11}{2}\times \frac{1}{17}\times \frac{1}{1}\times \frac{1}{3}$$
>>>>>$$\ \ \ = \frac{11}{102}$$
This is the probability of _not picking any women_. Thus, the probability of picking _at least one woman_ is
>$$\text{Probability} = 1-\frac{11}{102} = \frac{91}{102}$$.
Incorrect.
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Note that this is the probability of __not picking any women__!
Incorrect.
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Note that this is the probability of picking _exactly_ one woman, not of picking _at least_ one woman.
Incorrect.
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Incorrect.
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$$\frac{11}{102} $$
$$\frac{77}{204}$$
$$\frac{77}{102}$$
$$\frac{91}{102}$$
$$\frac{31}{34}$$
