Question 265805: Eight students are chosen at random from a group of twenty of which Alex, Beth and Fraser are a part. What is the probability that Alex, Beth and Fraser are all among the chosen?
I have tried the following:
8C3/18C10 = 0.00128
which seems WAY too small a probability.
Any thoughts?
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Eight students are chosen at random from a group of twenty of which Alex, Beth and Fraser are a part. What is the probability that Alex, Beth and Fraser are all among the chosen?
I have tried the following:
8C3/18C10 = 0.00128
which seems WAY too small a probability.
Any thoughts?
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P(A,B,C are in the group chosen) = 1 - P(they are not among the chosen)
---
P(they are not among the chosen) = 17C8/20C8 = 24310/125970 = 0.1930..
---
So P(A,B,C are in the group chosen = 1-0.1930 = 0.8070
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Cheers,
Stan H.
Answer by Edwin McCravy(20062)  (Show Source):
You can put this solution on YOUR website! Eight students are chosen at random from a group of twenty of which Alex, Beth and Fraser are a part. What is the probability that Alex, Beth and Fraser are all among the chosen?
8C3/18C10 = 0.00128
which seems WAY too small a probability.
Yes that's entirely wrong. The other tutor's answer is wrong
too. He found the probability that at least 1 of the three
got in the chosen group of 8. But that was not what you wanted.
You wanted the probability that all three were chosen, not
just at least one of them. The probability he got is
high, because it's pretty likely that at least one of them
was chosen, but it's not very likely that all three were chosen.
Here's the correct way to work it:
To be successful you have to choose
those three and 5 others. There is 1 way to choose those three
(namely to chose them!) and 17C5 ways to choose the other 5 from
the 17 besides Alex, Beth and Fraser.
Calculate the number of successful ways:
That's 1 way to choose the three --- TIMES 17C5 ways to choose
the other five, or 1x17C5 or 17C5
Now calculate the number of possible ways:
That's 20C8
Edwin
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