Question 149493: In a shop there are 20 customers, 18 of whom will make a purchase. If three customers are selected, one at a time, at random, what is the probability that all will make a purchase? Is it 0.7717, 0.7605, 0.8524, or 0.8808.
Please help
Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! In a shop there are 20 customers, 18 of whom will make a purchase. If three customers are selected, one at a time, at random, what is the probability that all will make a purchase? Is it 0.7717, 0.7605, 0.8524, or 0.8808.
Please help
It's not any of those. It's 0.7158
P(1st chosen will purchase
AND
2nd chosen will purchase
AND
3rd chosen will purchase)
=
P(1st chosen will purchase)
TIMES
P(2nd chosen will purchase, given that the 1st chosen will purchase)
TIMES
P(3rd chosen will purchase, given that the 1st and 2nd chosen will purchase)
=
 or rounded to ten-thousandths,
 .
You can also do it using combinations:
=
Stanbon's solution is incorrect because
the choices are not independent. Binomial
probabilities are ONLY for independent choices.
The choices are not independent because the
probability of selecting the second one as a
purchaser CHANGES after selecting the first one.
Also the probability of selecting the third one
as a purchaser CHANGES after selecting the
first two.
Edwin
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! In a shop there are 20 customers, 18 of whom will make a purchase.
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P(a random person in the group makes a purchase) = 18/20 = 9/10
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If three customers are selected, one at a time, at random, what is the probability that all will make a purchase?
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This is a binomial problem with p=0.9
P(3 of 3 make a purchase) = 0.9^3 = 0.729
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Cheers,
Stan H.
Is it 0.7717, 0.7605, 0.8524, or 0.8808.
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