# Help with a simple probability problem

• GeoMike
In summary, the probability of selecting exactly 8 women in a group of 20 motorcycle drivers is (20!/(12!*8!)) * (1/20)^8(19/20)^12 and the probability of selecting at least 8 women is the sum of the probabilities of exactly 8 through exactly 20 women.
GeoMike
Ok, this seems like an easy enough problem, but I'm stumped...

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If a motorcycle driver is selected at random from the US population, the probability of that driver being a male is 0.95
If 20 motorcycle drivers are selected at random from the population, what is the probability that exactly 8 will be women? What is the probability that at least 8 will be women?
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I've got that the chance of selecting a female is 1/20. And because the population is so large I can treat the events as independent.

I can easily find the probability of selecting 8 women in a row. What's throwing me off is the fact that there are 20 selections, so I know that the probability of selecting 8 total isn't going to be the same as selecting 8 in a row. I can't figure out how to account for the fact that 20 selections increases the probability of 8 women being selected.
I'm also thrown off by the at least 8 vs. exactly 8 part of the problem.

Any hints would be great. Thanks!
-GeoMike-

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if event A has a probability p of occurring, and an event B has a probability 1-p of occurring, and one event or the other occurs n times, you can characterize the probability by

pk(1-p)n-k

where k is the number of times A happens (this assumes order counts. In your problem it doesn't, so you'll need to fix it for that)

Ok, that makes sense. So, the probability of getting 8 females and 12 males (in a specific combination) would be:

(1/20)8(19/20)12

To account for the fact that any arrangement of 8 females and 12 males is acceptable, I'd just multiply that probability by the total number of combinations:

(20!/(12!*8!)) * (1/20)8(19/20)12 = P(exactly 8 women)

Right?

And to find P(at least 8 women) I'd just sum the probabilities of exactly 8 through exactly 20 women:

$$\sum_{i=8}^{20}{\frac{20!}{(20-i)!i!}(1/20)^{x}(19/20)^{20-x}}$$

Right?

Thanks,
-GeoMike-

of course, you'll have fewer things to sum of you do 0 to 7 females... and something about 1-p

Although, in this case, it's not that big a deal. But, if it was "at least 2 females", it'd be far quicker to find the probabilities of 0 females and 1 female, then subtracting from 1.