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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?

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?

0.343 0.147 0.189 0.063 0.027

I am wondering why C is the correct answer. I keep solving .3*.3*.7 which is .063 but in the reason why its correct it says that I should multiply by 3!/2! which doesnt make sense. Can anyone help. Thanks

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?

0.343 0.147 0.189 0.063 0.027

I am wondering why C is the correct answer. I keep solving .3*.3*.7 which is .063 but in the reason why its correct it says that I should multiply by 3!/2! which doesnt make sense. Can anyone help. Thanks

you need to multiply 0.063 by 3 because there 3 ways 2 visitors can be chosen;

Possible scenarios: C C X C X C X C C Case I: 0.3x0.3x0.7 Probabilities will be added for all 3 scenarios(prob for each scenario is the same .3x.3x.7) Total probability: 3x(.3x.3x.7) =.189 Alternate: 3C2 : selecting 2 out of 3 where order isn't important C1, C2 cant be differentiated. Prob: 3C2 x .3x.3x.7 =.189

would somone please post a reply to ra011's question. I would really like to know the answer, i get stumped by almost all of the probability questions as there always seems to be some new reason why we should do it differently. Please help.

If the question were to state : "what is the probability that any two will buy a pack of candy?" then would the answer be (0.3 x 0.3 x 0.7) = 0.063?

Am still trying to get my head wrapped around this probability thingy.

Thanks!!

Think about it, if the constraint says "any two", it is less restrictive than saying a specific 2 out of the 3 or even exactly 2 out of 3. So any two out of the three buying candy would be the same answer as the OA + the probability of all 3 buying candy. (Since even in this case, 2 of them do buy candy, we have no restriction on the 3rd)

i.e. 0.189 + 0.3*0.3*0.3 = 0.216

Correct me if I'm wrong though.

EDIT: 0.063 would have been the answer in the following case: A, B and C visit the mall. What is the probability that only A and B will buy candy, and C does not?

A lot of people always make this same mistake. When it comes to probability combined with combination problems, you have to consider the no of combination it can be arranged in.

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?