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I've placed one A, two B's, and four C's into a hat.

What is the probability of pulling two C's out of the hat with replacement? 
 
What is the probability of pulling two C's out of the hat without replacement? 
 
What is the probability of pulling two A's out of the hat without replacement? 
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1 Answer

1 A, 2 B's, 4 C's
 
probability of 2 C's with replacement:
(4/7)(4/7)=16/49
probability of 2 C's without replacement:
(4/7)(3/6)=(4/7)(1/2)=(4/14)=2/7
probability of 2 A's:
0 because: (1/7)(0/6)=(1/7)(0)=0, in other words it's not possible without replacement because you only have one A

Comments

Great answer! I just wanted to add, in case it's not clear to the student, that "with replacement" means you're putting the A, B, or C you draw back into the hat before drawing from the hat again. "Without replacement" means that, once you draw an A, B, or C from the hat, you don't put it back into the hat.
Thanks Stephanie,
                         Good point about the word "replacement". Students sometimes use words they see without really knowing what they mean.
Arthur D.