SOLUTION: A bag contains 20 M&M's; 5 are blue, 9 are red, and 6 are yellow
a) If 2 M&M's are randomly selected without replacement, what is the probability both are blue?
For this, I tried
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-> SOLUTION: A bag contains 20 M&M's; 5 are blue, 9 are red, and 6 are yellow
a) If 2 M&M's are randomly selected without replacement, what is the probability both are blue?
For this, I tried
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Question 275485: A bag contains 20 M&M's; 5 are blue, 9 are red, and 6 are yellow
a) If 2 M&M's are randomly selected without replacement, what is the probability both are blue?
For this, I tried:
P (2 blue) = (5/20)(4/19)
b) If 2 M&Ms are randomly selected without replacement, what is the probability at least 1 is blue?
For this, i tried:
P (at least 1 is blue)
1- P (none blue) = 1-(15/20) Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! nCr=combination: n things taken r at a time=n!/((n-r)!r!)
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a)
5C2/20C2
=10/190
=1/19 probability both are blue.
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b)
15C2/20C2 probability none are blue
=105/190
=21/38
1- 21/38 = 17/38 probability at least 1 is blue.
.
Ed
