A snack-size bag of M&Ms candies is opened. Inside, there are 12 red candies, 12 blue, 7 green, 13 brown, 3 orange, and 10 yellow. Three candies are pulled from the bag in succession, without replacement.
What is the probability that the first two candies drawn are orange and the third is green?
First, we need to calculate the total number of candies in the bag:
12 (red) + 12 (blue) + 7 (green) + 13 (brown) + 3 (orange) + 10 (yellow) = 57 candies
Now we can calculate the probability of pulling an orange candy on the first draw:
P(orange on first draw) = 3/57 = 1/19
Then, since the candy is not replaced, the probability of pulling another orange candy on the second draw is:
P(orange on second draw) = 2/56 = 1/28
Finally, the probability of pulling a green candy on the third draw is:
P(green on third draw) = 7/55
Now we multiply the individual probabilities together to find the total probability:
P(orange, orange, green) = P(orange on first draw) * P(orange on second draw) * P(green on third draw)
P(orange, orange, green) = (1/19) * (1/28) * (7/55)
P(orange, orange, green) ≈ 0.00070
Therefore, the probability that the first two candies drawn are orange and the third is green is approximately 0.00070, or 0.07%.
i need the answer as a fraction.
To find the probability as a fraction, we first calculate the individual probabilities as fractions:
P(orange on first draw) = 3/57 = 1/19
P(orange on second draw) = 2/56 = 1/28
P(green on third draw) = 7/55
Now we multiply the fractions together to find the total probability:
P(orange, orange, green) = (1/19) * (1/28) * (7/55)
P(orange, orange, green) = 7/29260
Therefore, the probability that the first two candies drawn are orange and the third is green is 7/29260.
youre incorrect. the answer is 1/4180
Apologies for the mistake. Let's calculate the probability again.
First, we need to calculate the total number of candies in the bag:
12 (red) + 12 (blue) + 7 (green) + 13 (brown) + 3 (orange) + 10 (yellow) = 57 candies
Now we can calculate the probability of pulling an orange candy on the first draw:
P(orange on first draw) = 3/57 = 1/19
Then, since the candy is not replaced, the probability of pulling another orange candy on the second draw is:
P(orange on second draw) = 2/56 = 1/28
Finally, the probability of pulling a green candy on the third draw is:
P(green on third draw) = 7/55
Now we multiply the individual probabilities together to find the total probability:
P(orange, orange, green) = P(orange on first draw) * P(orange on second draw) * P(green on third draw)
P(orange, orange, green) = (1/19) * (1/28) * (7/55)
P(orange, orange, green) = 1/4180
Therefore, the correct probability that the first two candies drawn are orange and the third is green is 1/4180. Thank you for pointing out the error.