Question

Chelsey puts twelve colored pieces of chocolate candy in a bag. Three pieces are green, four pieces are yellow, two pieces are purple, and three pieces are red. She draws one piece of candy from the bag at a time and eats it.

A. What is the probability that Chelsey will draw out three pieces of candy in this order: red, green, yellow?
B. What is the probability that Chelsey will draw out three pieces of candy in this order: green, yellow, red?
C. What is the probability that Chelsey will draw out three pieces of candy in this order: yellow, red, green? D. Make a conclusion about how the order in which she draws out the red, green, and yellow pieces affects the probability.

Answer 1

The probability of drawing three candies in a specific order can be calculated by multiplying the probabilities of each event happening in sequence. The order in which the candies are drawn affects the probability because the probability of each event is dependent on the previous events.

Step-by-step explanation:

A. To find the probability of drawing three pieces of candy in the order red, green, yellow, we need to find the probability of each event happening in sequence. The probability of drawing a red candy first is 3/12, since there are 3 red candies out of 12 total candies. After drawing a red candy, there are 2 red candies left out of 11 total candies. Therefore, the probability of drawing a green candy next is 2/11. Finally, after drawing a red and green candy, there are 4 yellow candies left out of 10 total candies. So the probability of drawing a yellow candy last is 4/10. To find the overall probability, we multiply these individual probabilities together: (3/12) * (2/11) * (4/10) = 24/1320 = 1/55.

B. To find the probability of drawing three pieces of candy in the order green, yellow, red, we follow the same process. The probability of drawing a green candy first is 4/12, since there are 4 green candies out of 12 total candies. After drawing a green candy, there are 3 green candies left out of 11 total candies. Therefore, the probability of drawing a yellow candy next is 3/11. Finally, after drawing a green and yellow candy, there are 3 red candies left out of 10 total candies. So the probability of drawing a red candy last is 3/10. Multiplying these probabilities together gives us: (4/12) * (3/11) * (3/10) = 36/1320 = 3/110.

C. To find the probability of drawing three pieces of candy in the order yellow, red, green, we follow the same process. The probability of drawing a yellow candy first is 2/12, since there are 2 yellow candies out of 12 total candies. After drawing a yellow candy, there are 3 yellow candies left out of 11 total candies. Therefore, the probability of drawing a red candy next is 3/11. Finally, after drawing a yellow and red candy, there are 3 green candies left out of 10 total candies. So the probability of drawing a green candy last is 3/10. Multiplying these probabilities together gives us: (2/12) * (3/11) * (3/10) = 18/1320 = 3/220.

D. The order in which the candies are drawn affects the probability because the probability of each event is dependent on the previous events. If we change the order of the candies to be drawn, the probabilities of each event happening will be different. For example, if Chelsey drew the candies in a different order, the probabilities in parts A, B, and C would be different.