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A store sells 3 items, the chance that a customer will buy item A is 45%, item B 15% and item C is 40%. If 5 customers come to the store, what is the probability that 2 item A’s and 1 item B will be purchased?

My attempt is to multiply the chances of each item that is purchased and the number of customers so

0.45*0.45*0.15*5 pretty sure it’s incorrect though. Please someone explain it to me

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    $\begingroup$ Is each customer buying only one item? $\endgroup$ – Anurag A Oct 8 '19 at 3:43
  • $\begingroup$ It is not given in the question $\endgroup$ – SVS Oct 8 '19 at 3:47
  • $\begingroup$ @AnuragA Since the probabilities add up to $100\%$, I think we can take it that each customer buys exactly one item, though I admit it's ambiguous. It might mean that the probability that he buys at least one A is $45\%$, but then I don't think it's possible to answer the question. $\endgroup$ – saulspatz Oct 8 '19 at 3:49
  • $\begingroup$ You are ignoring a couple of things. First, the other two customers must but C. Second, we have choices as to which customers buy A and which customer buys B. $\endgroup$ – saulspatz Oct 8 '19 at 4:11
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The question is a little bit ambiguous. I am assuming that it is asking us for the probability of the customers EXACTLY purchasing 2 A items and 1 B item and not the probability of AT LEAST 2 As and 1 B. Lets look at the event of interest. We want 5 customers to select exactly 2As and 1B so the only way this is satisfied is if the other two remaining select C. So we want:

$$P(A)*P(A)*P(B)*P(C)*P(C)=.45^2*.15*.4^2=.00486$$

So that exactly 2 A items are selected and 1 B item.

BUT there are 30 different ways that our event of interest can happen. I will denote the customer with a number, followed by the item he/she picked.

Note that one of our events of interest: $$1C,2C,3A,4A,5C$$ is not the same as the event say: $$1C,2C,3B,4A,5A$$

Note that for every different way of picking our two C customers, there are 3 ways to pick items A and B: $$1A,2A,3B$$ $$1A,2B,3A$$ $$1B,2A,3A$$

and we can take two C items in ${5\choose 2}=10$ different ways.

For each one of them there are 3 possible combinations of A and B items that interest us. So in total, our event of interest:

2 items A and 1 B has 30 different ways of happening and the probability that it happens is $.00486$. Which gives us:

P(exactly 2As and 1B)$=.00486*30=0.1458$

So our desired probability is 14.58%

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