Probability without Replacement

Question

Dr. Emmett Brown has a bag of M&Ms. He has 7 green M&Ms, 12 brown M&Ms, 3 blue M&Ms, and 5 red M&Ms.

What is the probability that Emmett selects a blue M&M followed by a green M&M without replacement?

Please use four decimals.

Thank you so much

Answer 1

3*7/(27*26) = 7/234 = .0030

Answer 2

If you add all the m&m's in the bag, there are a total of 27. If you are sampling without replacement, you will take 1 out of the bag and there will only be 26 m&m's left in the bag for the second draw.

The probability that the first one will be blue is 3/27 since there are 3 blue in the bag.

Assuming we drew a blue the first time, the probability that the second one will be green is 7/26 since drawing a blue doesn't change the number of green in the bag but does change the total in the bag.

The probability of both events is the product of the individual probabilities: 3/27 x 7/26 = 21/702 ≈ 0.0299