Question

1. A fun size bag of M&Ms has 4 blue, 3 orange, 3 red, 2 green, 2 yellow, and 1 brown M&Ms. What is the probability of randomly selecting 5 M&Ms where 3 are blue and 2 are red?

Answer 1

Solution:
Given in the question
Fun size bag has 4 blue, 3 orange, 3 red, 2 green, 2 yellow and 1 brown M&Ms
So total No. of M&Ms are = 4+3+3+2+2+1 = 15
No. of randomly selected M&Ms are 5
Out of 5 M&Ms, 3 are blue and 2 are red so
Total no. of ways 3 blue M&Ms out of 4 = 4C3 = 4
Total no. of ways 2 red M&Ms out of 3 = 3C2 = 3
So total no. of ways 3 are blue and 2 are red = Total no. of ways 3 blue M&Ms out of 4 * Total no. of ways 2 red M&Ms out of 3 = 4*3 = 12
Total no. of ways 5 M&Ms out of toal 15 = 15C5 = (15!/(5!*(15-5)!) = 3003

probability of randomly selecting 5 M&Ms where 3 are blue and 2 are red = (Total no. of ways 3 blue M&Ms out of 4 * Total no. of ways 2 red M&Ms out of 3)/Total no. of ways 5 M&Ms out of toal 15 = 4*3/3003 = 12/3003 = 0.004