Araceli has a bag containing plain chocolate pieces and another bag containing peanut chocolate pieces. The bag with plain chocolate pieces contains 5 blue, 1 yellow, 2 green, and 4 red candies. The bag with peanut chocolate pieces contains 3 blue, 1 yellow, and 4 green candies. Araceli will randomly select 1 candy from each bag.What is the probability that Araceli will select a blue or yellow candy from each bag?ResponsesA 343 4B 141 4C 121 2D 110
Question
Araceli has a bag containing plain chocolate pieces and another bag containing peanut chocolate pieces. The bag with plain chocolate pieces contains 5 blue, 1 yellow, 2 green, and 4 red candies. The bag with peanut chocolate pieces contains 3 blue, 1 yellow, and 4 green candies. Araceli will randomly select 1 candy from each bag.What is the probability that Araceli will select a blue or yellow candy from each bag?ResponsesA 343 4B 141 4C 121 2D 110
Solution
To solve this problem, we first need to calculate the total number of candies in each bag and the number of favorable outcomes (blue or yellow candies) in each bag.
In the plain chocolate bag, there are 5 blue + 1 yellow + 2 green + 4 red = 12 candies in total. Out of these, 5 blue + 1 yellow = 6 are favorable outcomes.
In the peanut chocolate bag, there are 3 blue + 1 yellow + 4 green = 8 candies in total. Out of these, 3 blue + 1 yellow = 4 are favorable outcomes.
The probability of selecting a blue or yellow candy from the plain chocolate bag is 6/12 = 1/2.
The probability of selecting a blue or yellow candy from the peanut chocolate bag is 4/8 = 1/2.
Since these are independent events, we multiply the probabilities to get the overall probability. So, the probability that Araceli will select a blue or yellow candy from each bag is 1/2 * 1/2 = 1/4.
So, the correct answer is B. 1/4.
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