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Probability

by shashwats » Thu May 02, 2013 12:13 pm
If there are 10 balls, 3 red, 5 blue and 2 green. I pick two without replacement, what is the probability that both are red?

I find two ways of doing this, clearly one is wrong!

1) 3/10 * 2/9

2) (3C1 *2C1)/10C2

Can someone please help me understand this!
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by srcc25anu » Thu May 02, 2013 12:48 pm
Prob of 1st red ball = 3/10
Prob of 2nd red ball = 2/9
Prob of 2 red balls = 3/10 * 2/9 = 1/15

Alternate approach:
Ways to select 2 red balls out of total 3 red balls = 3C2 = 3
Ways to select any 2 balls out of total 10 balls = 10C2 = 45
Prob of 2 red balls from total 2-ball selections = 3/45 or 1/15
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by srcc25anu » Thu May 02, 2013 12:56 pm
shashwats wrote: 2) (3C1 *2C1) / 10C2
When you write 3C1 *2C1, it means first you select 1 out of 3 red balls (r1, r2 and r3) in 3C1 = 3 ways. and then you select 1 more ball in remaining 2 red balls in 2 ways. if you selected first ball r1, now you can select r2 or r3. SO total combinations here becomes 3 * 2 = 6 which is wrong.
Total combinations here will be {(r1),(r2)}, {(r1),(r3)},{(r2),(r3)},{(r3),(r2)},{(r3),(r1)},{(r2),(r1)}. (r1)(r2) combination is different from (r2)(r1) combination in this case.

we have to consider all the 2-red ball combinations together which can be (r1,r2), (r1,r3) and (r2,r3). There is No Other Combination.

NUTSHELL: 3C2 is Not the same as 3C1 * 2C1
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by Brent@GMATPrepNow » Fri May 03, 2013 4:26 am
shashwats wrote:If there are 10 balls, 3 red, 5 blue and 2 green. I pick two without replacement, what is the probability that both are red?
P(both red) = P(1st is red AND 2nd is red)
= P(1st is red) x P(2nd is red)
= 3/10 x 2/9
= 6/90
[spoiler]= 1/15[/spoiler]

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