fortsill wrote:

where's the right answer posted? how do i know if i'm right?

I added the correct answer under the spoiler in the initial post. Below is a solution.

The probability that a visitor at the mall buys a pack of candy is 0.3. If 6 visitors came to the mall today, what is the probability that exactly 4 will buy candy?Since the probability that a visitor buys a candy is 0.3, then the probability that a visitor DOES NOT buy a candy is 1-0.3=0.7.

We want the probability that exactly 4 visitors out of 6 will buy a candy, so the probability of BBBBNN (where B denotes a visitor who buys a candy and N denotes a visitor who does not buy a candy). Each B has the probability of 0.3 and each N has the probability of 0.7, so we have \(0.3^4*0.7^2\).

Next, BBBBNN case can occur in # of different ways: NNBBBB, NBNBBB, NBBNBB, ... (first two visitors doesn't buy and next four does; first doesn't buy, second does, third doesn't and next three does; ...) Basically it's # of permutations of 6 letters BBBBNN out of which 4 B's and 2 N's are identical, so \(\frac{6!}{4!*2!}\).

Finally \(P(B=4)=\frac{6!}{4!*2!}*0.3^4*0.7^2\).

Check the following links for different scenarios similar problems:

the-probability-that-a-visitor-at-the-mall-buys-a-pack-of-55689.htmlprobability-85523.htmlHope it helps.

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