Question

a box has 12 candies in it: 3 are butterscotch, 4 are caramel, and 5 are peppermint. (each candy falls into only one of these categories.) boris 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 two candies selected are caramel? do not round your intermediate computations. round your final answer to three decimal places.

Explanation:

Step1: Probability first caramel

There are 4 caramel candies out of 12 total. So the probability of selecting a caramel first is $\frac{4}{12}$.

Step2: Probability second caramel

After selecting one caramel, there are 3 caramel left and 11 total candies. So the probability of selecting a caramel second is $\frac{3}{11}$.

Step3: Multiply probabilities

To find the probability of both events happening (first caramel and second caramel), we multiply the two probabilities: $\frac{4}{12} \times \frac{3}{11} = \frac{12}{132} = \frac{1}{11} \approx 0.090909...$

Step4: Round to three decimals

Rounding 0.090909... to three decimal places gives 0.091.

Answer:

0.091