Here we go, the question stem asks about EXACT number of events. Hence, to get the exactly the probability of two visitors buy a pack of candy can be find using binomial probability, or Bernulli trials consisting following steps:

1. 3C2 - number of ways how exactly 2 visitors could buy a pack of candy. Here it is easy, 3 (like this A,B,C visitors and we have AB, AC and BC - 3 ways).

2. raising our beneficial probability, i.e. 30% in our case to the power ow 2-exactly 2 customers in our case, hence 0.3^2 = 0.09

3. raising our failure probability, i.e. 1-0.3=0.7 to the power of 3-2=1, failure non-beneficial events.

Eventually, we just multiply all aforesaid three values and get the required probability, P = 3*0.09*0.7=27*7/1000 = 189/1000 or 0.189

So, I bet the answer should be (C), according to the Bernulli trials, I just followed what that math genius stated.

Please, correct me if I went awry.

georgechanhc wrote:

The probability that a visitor at the mall buys a pack of candy is 30%. If three visitors come to the mall today, what is the probability that exactly two will buy a pack of candy?

(A) 0.343

(B) 0.147

(C) 0.189

(D) 0.063

(E) 0.027

Source: GMAT Club Tests - hardest GMAT questions

The explained answer is:

0.3 x 0.3 x 0.7 x 2C3 (combination) = 0.3 x 0.3 x 0.7 x 3 = 0.189

Why is it not just 0.3 x 0.3 x 0.7? What is the significance of 2C3?

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