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a box has 10 candies: 3 are peppermint, 2 are butterscotch, and 5 are t…

Question

a box has 10 candies: 3 are peppermint, 2 are butterscotch, and 5 are taffy. (each candy falls into only one of these categories.) lisa 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 taffy? do not round your intermediate computations. round your final answer to three decimal places. (if necessary, consult a list of formulas.)

Explanation:

Step1: Calculate total initial candies

Total candies = $2 + 3 + 10 = 15$

Step2: Prob first is butterscotch

Probability (first = butterscotch) = $\frac{3}{15}$

Step3: Adjust total for second pick

Remaining candies = $15 - 1 = 14$

Step4: Prob second is taffy

Probability (second = taffy) = $\frac{10}{14}$

Step5: Multiply the two probabilities

Combined probability = $\frac{3}{15} \times \frac{10}{14}$
Simplify: $\frac{30}{210} = \frac{1}{7} \approx 0.142857$

Step6: Round to 3 decimal places

Round $0.142857$ to 3 decimals: $0.143$

Answer:

0.143