bblast 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 visitors will buy a pack of candy?

0.343
0.147
0.189
0.063
0.027

C

[spoiler]before marking D for this I gave a thought that it should be C

(a,b not c)+(a,c not b) + (b,c not a) = .63+.63+.63. However, I was not sure.
How to decide and mark it in my mind ? I do get mixed up at times despite knowing how to solve any probablilty P&C question.[/spoiler]

Your reasoning is correct: "exactly" type probability questions must account for the order of choosing the events. Exactly two candies could be 1st and 2nd, 1st and 3rd, 2nd and 3rd.

Generally speaking for this (rare) type of questions, the scenarios should be symetrical - each of the three scenarios above will have the same equal probability. Therefore, it is enough to find the probability of a single scenario (say, A, B, not C), and multiply that multiply that fraction by the number of scenarios - in this case, 0.063*3.

bblast 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 visitors will buy a pack of candy?

0.343
0.147
0.189
0.063
0.027

C

[spoiler]before marking D for this I gave a thought that it should be C

(a,b not c)+(a,c not b) + (b,c not a) = .63+.63+.63. However, I was not sure.
How to decide and mark it in my mind ? I do get mixed up at times despite knowing how to solve any probablilty P&C question.[/spoiler]

Probability that exactly two visitors out of three will buy a pack of candy

= 3 (choose) 2 × 0.3 × 0.3 × 0.7

= 3 × 0.063

= [spoiler]0.189


C[/spoiler]