Question

A box has 12 candies. 6 are butterscotch, 3 are taffy, and 3 are peppermint (each candy falls into only one of these categories). Bob wants to select two candies to eat for dessert. The first candy will be selected at random, and then the second candy will be selected at random from the remaining candies. What is the probability that the first candy selected is butterscotch and the second candy is peppermint?

Answer 1

3/22

Explanation:

To find the probability that the first candy selected is butterscotch and the second candy is peppermint, we can break it down into two steps:

Step 1: Probability of selecting a butterscotch candy first.

There are 12 candies in total, and 6 of them are butterscotch. So, the probability of selecting a butterscotch candy first is:

Probability (butterscotch first) = (Number of butterscotch candies) / (Total number of candies)

Probability (butterscotch first) = 6 / 12

Step 2: Probability of selecting a peppermint candy second.

After the first candy is selected, there are now 11 candies left in the box, and 3 of them are peppermint (as one butterscotch candy has already been chosen). So, the probability of selecting a peppermint candy second is:

Probability (peppermint second) = (Number of peppermint candies) / (Total number of remaining candies)

Probability (peppermint second) = 3 / 11

Now, to find the overall probability of both events happening in sequence (butterscotch first and peppermint second), you multiply the probabilities from each step:

Probability (butterscotch first and peppermint second) = (Probability butterscotch first) * (Probability peppermint second)

Probability (butterscotch first and peppermint second) = (6 / 12) * (3 / 11)

Now, calculate this probability:

Probability (butterscotch first and peppermint second) = (1/2) * (3/11) = 3/22

So, the probability that the first candy selected is butterscotch, and the second candy is peppermint is 3/22.