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)
4 red chips and 2 blue chips
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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 :idea:](./images/smilies/Idea.png)
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
![Wink ;)](./images/smilies/wink.png)
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 :idea:](./images/smilies/Idea.png)
-
- Master | Next Rank: 500 Posts
- Posts: 447
- Joined: Sun Apr 19, 2009 9:08 pm
- Location: Kolkata,India
- Thanked: 7 times
- GMAT Score:670