Question

a box has 12 candies in it: 6 are taffy, 3 are butterscotch, and 3 are peppermint. (each candy falls into only one of these categories.) eric 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?

do not round your intermediate computations. round your final answer to three decimal places.

(if necessary, consult a list of formulas.)

Explanation:

Identify the given values and target event

We are given a box containing a total of \(12\) candies:

• Taffy: \(6\)
• Butterscotch: \(3\)
• Peppermint: \(3\)

Eric selects two candies sequentially without replacement. We want to find the probability that the first candy selected is butterscotch (\(B_1\)) and the second candy selected is peppermint (\(P_2\)).

Calculate the probability of the first event

Using the Probability of Events concept, we find the probability of selecting a butterscotch candy first. Since there are \(3\) butterscotch candies out of \(12\) total candies:

$$ P(B_1) = \frac{3}{12} = \frac{1}{4} $$

Calculate the conditional probability of the second event

Since the first candy is not replaced, there are now \(11\) candies remaining in the box. The number of peppermint candies remains \(3\).
The probability of selecting a peppermint candy second, given that a butterscotch was selected first, is:

$$ P(P_2 \mid B_1) = \frac{3}{11} $$

Calculate the joint probability

To find the probability of both events occurring, we multiply the probability of the first event by the conditional probability of the second event:

$$ P(B_1 \cap P_2) = P(B_1) \times P(P_2 \mid B_1) = \frac{3}{12} \times \frac{3}{11} = \frac{9}{132} $$

Simplify and round the final answer

We simplify the fraction and convert it to a decimal:

$$ P(B_1 \cap P_2) = \frac{3}{44} \approx 0.0681818... $$

Rounding to three decimal places gives \(0.068\).

Answer:

A box has 12 candies in it: 6 are taffy, 3 are butterscotch, and 3 are peppermint. (Each candy falls into only one of these categories.) Eric 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?

Do not round your intermediate computations. Round your final answer to three decimal places.

<blank>0.068</blank>