4 red chips and 2 blue chips

[uptowngirl92]

Guys please tell me where am i going wrong:

A box contains 4 red chips and 2 blue chips. If two chips are selected at random without replacement, what is the probability that the chips are different colors?


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My solution:
Got the correct answer by the following method.
(4/6)(2/5)+(2/6)(4/5)= 8/15>>OA

My querie:
Why are we not getting the answer when we do 1-same color??
1-Both red + 1-both blue
(4/6)(3/6)+(2/6)(1/6)

[NikolayZ]

Hey Uptowngirl!
Actually we can do this.
In 1-x shortcut, (under the right solution) you forgot that we are not replacing chips.
So.
4/6*3/5+2/6*1/5=7/15 1-7/15=8/15.
So this approach is alright!

Also you could do this problem with combinatorics.
Total outcomes : 6!/4!2!=15, then we taking 1 out of 4 red chips , 4!/3!=4, and 1 out of 2 blue chips. 2!/1!=2.
4*2/15=8/15

[uptowngirl92]

thank you..I'll go jump off the balcony now..!!grumble grumble..Silly silly!!