SOLUTION: A box contains a total of 12 crayons: 2 red, 3 green, 3 blue, 1 yellow, 2 purple, and 1 brown. Without looking, Frieda picks two crayons from the box. What is the probability that

Algebra ->  Finance -> SOLUTION: A box contains a total of 12 crayons: 2 red, 3 green, 3 blue, 1 yellow, 2 purple, and 1 brown. Without looking, Frieda picks two crayons from the box. What is the probability that       Log On


   

Question 1112807: A box contains a total of 12 crayons: 2 red, 3 green, 3 blue, 1 yellow, 2 purple, and 1 brown. Without looking, Frieda picks two crayons from the box. What is the probability that BOTH will be blue?
A.5/23
B.1/22
C.1/4
D.19/44
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
ONE WAY TO LOOK AT THE SITUATION:
As Frieda wraps her hand around one crayon
out of the 12 in the box,
the probability that she is grabbing one of the 3 blue ones is
3%2F12=1%2F4 .
IF/after she gets grabs a first blue crayon,
as she gropes for a second crayon from the 12-1=11 remaining,
the probability that she gets one of the other 3-1=2 blue crayons is
2%2F11.
The probability of those two grabs is
%281%2F4%29%282%2F11%29=highlight%281%2F22%29 .

ANOTHER WAY (looking at combinations):
It the 12 crayons were marked to be able to keep track of each one individually,
we would figure out that there are
12%2A11%2F2=66 different sets of two crayons that Frieda could pick up,
while there are 3 different sets of two blue crayons that could be picked up.
The pairs of blue crayons as a fraction of all pairs of crayons,
3%2F66=highlight%281%2F22%29 is the probability that both crayons picked by Frieda are blue.