Asked by Todd
A bag of M&Ms contains 12 red, 11 yellow, 5 green, 6 orange, 5 blue, and 16 brown candies. What is the probability that if you choose 2 M&Ms from the bag (one after the other) without looking, you will choose 2 yellow ones?
Answers
Answered by
MathMate
First calculate the total numebr of candies:
12+11+5+6+5+16=55
(check my work, because my calculator sometimes misses a digit).
There are 11 yellow candies out of the 55.
If you pick (not choose) without looking <i>one after another</i>, this is what is called <i>pick without replacement</i>.
If two yellow candies are required, the probability to pick the first one is therefore 11/55.
Since there are now 10 left out of 54 (without replacement), the probability of picking a yellow <i>this time</i> is 10/54.
The probability that <i>both</i> events will occur is therefore the product of the two, namely
11/55*10/54.
I will leave you to calculate the final answer. Don't forget to cancel common factors if you present it in fraction form (my preference).
12+11+5+6+5+16=55
(check my work, because my calculator sometimes misses a digit).
There are 11 yellow candies out of the 55.
If you pick (not choose) without looking <i>one after another</i>, this is what is called <i>pick without replacement</i>.
If two yellow candies are required, the probability to pick the first one is therefore 11/55.
Since there are now 10 left out of 54 (without replacement), the probability of picking a yellow <i>this time</i> is 10/54.
The probability that <i>both</i> events will occur is therefore the product of the two, namely
11/55*10/54.
I will leave you to calculate the final answer. Don't forget to cancel common factors if you present it in fraction form (my preference).
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