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4 red chips and 2 blue chips

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4 red chips and 2 blue chips

by uptowngirl92 » Thu Oct 29, 2009 1:15 am
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?


............................................................
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)
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by NikolayZ » Thu Oct 29, 2009 1:29 am
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
:idea:
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Master | Next Rank: 500 Posts
Posts: 447
Joined: Sun Apr 19, 2009 9:08 pm
Location: Kolkata,India
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by uptowngirl92 » Thu Oct 29, 2009 2:25 am
thank you..I'll go jump off the balcony now..!!grumble grumble..Silly silly!!
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