Probability: If there are 10 people, and 3 are randomly chosen,...

thunc14

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I think I may be overthinking this problem:

If there are 10 people, and 3 are randomly chosen, what is the probability that person A will be chosen and person B will not be chosen?

My confusion is if person A is the first, second, or third person chosen, how do I account for those different probabilities? At first person A's probability would be 1/10, then 1/9, then 1/8. I'm not sure how to account for this mathematically speaking, and how to account for person B not getting chosen.
 
I think I may be overthinking this problem:
If there are 10 people, and 3 are randomly chosen, what is the probability that person A will be chosen and person B will not be chosen?
My confusion is if person A is the first, second, or third person chosen, how do I account for those different probabilities? At first person A's probability would be 1/10, then 1/9, then 1/8. I'm not sure how to account for this mathematically speaking, and how to account for person B not getting chosen.
There are a total of \(\displaystyle \dbinom{10}{3}\) ways to select three of ten.
There are a total of \(\displaystyle \dbinom{8}{2}\) ways to select three that include A but not B.
Now what?
 
I understand (10 3) but I don't quite get how you arrived at (8 2). I'm thinking this is a permutation problem?
 
I understand (10 3) but I don't quite get how you arrived at (8 2). I'm thinking this is a permutation problem?
If we remove A & B from the group that leaves eight. If we choose two of those \(\displaystyle \binom{8}{2}=28\) ways none of which includes B.
Put A with any one of those then we have a subgroup of three that includes A but not B.

Selection problems are combinations if positions such as Pres. VP, & Sec. are not considered.
 
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If we remove A & B from the group that leaves eight. If we choose two of those \(\displaystyle \binom{8}{2}=28\) ways none of which includes B.
Put A with any one of those then we have a subgroup of three that includes A but not B.

Selection problems are combinations if positions such as Pres. VP, & Sec. are not considered.

I'm still a bit lost here. Why would we remove both A and B from the group, when we want person A to be selected. Still not sure how to calculate this probability.
 
I'm still a bit lost here. Why would we remove both A and B from the group, when we want person A to be selected. Still not sure how to calculate this probability.

You want to include A and not include B. So both of their statuses are already known. There are 8 other people to consider; you need to pick 2 of them, in addition to A, who is already chosen.

You might think of it this way: List the 10 people, under whom you are to mark 2 as chosen. We premark A as chosen and B as not:

A B C D E F G H I J
Y N _ _ _ _ _ _ _ _

So you have to pick 2 of the 8 under which to put Y, and you will have made the required choice.

Okay?
 
You want to include A and not include B. So both of their statuses are already known. There are 8 other people to consider; you need to pick 2 of them, in addition to A, who is already chosen.

You might think of it this way: List the 10 people, under whom you are to mark 2 as chosen. We premark A as chosen and B as not:

A B C D E F G H I J
Y N _ _ _ _ _ _ _ _

So you have to pick 2 of the 8 under which to put Y, and you will have made the required choice.

Okay?

Thank you for that explanation, I understand. So we have a total of (8 2) ways of picking the remaining two people, out of a total (10 3) ways of picking 3 people from the 10. Using the combination formula, the probability would then be (8C2 / 10C3), which equals 7/30, correct?
 
Thank you for that explanation, I understand. So we have a total of (8 2) ways of picking the remaining two people, out of a total (10 3) ways of picking 3 people from the 10. Using the combination formula, the probability would then be (8C2 / 10C3), which equals 7/30, correct?

Yes.
 
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