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probability

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by akhpad » Sun Aug 29, 2010 4:40 am
blaster wrote: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?

OA later

Method 1:
4C1 * 2C1 / 6C2 = 8/15


Method 2:
(4/6)(2/5)+(2/6)(4/5)= 8/15
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by diebeatsthegmat » Sun Aug 29, 2010 9:42 am
blaster wrote: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?

OA later
select 2 from 6 chips : 6!/2!4!=15
select 1 from 4 red chip: 4
select 1 from 2 blue chips: 2
thus totallly probability is 2*4/15=8/15
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Re: probability

by Brent@GMATPrepNow » Wed Feb 05, 2020 7:01 am
blaster wrote: ↑
Sun Aug 29, 2010 2:52 am
 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?

OA later
P(different colors) = P(1st chip is red and 2nd chip is blue OR 1st chip is blue and 2nd chip is red)
= P(1st chip is red and 2nd chip is blue) + P(1st chip is blue and 2nd chip is red)
= P(1st chip is red) x P(2nd chip is blue) + P(1st chip is blue) x P(2nd chip is red)
= 4/6 x 2/5 + 2/6 x 4/5
= 8/30 + 8/30
= 16/30
= 8/15

Answer: B

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Re: probability

by Scott@TargetTestPrep » Sun Feb 09, 2020 5:25 am
blaster wrote: ↑
Sun Aug 29, 2010 2:52 am
 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?

OA later
To satisfy the requirement, we must obtain either R-B or B-R. Thus, we need to determine:

P(red) x P(blue) + P(blue) x R(red)

4/6 x 2/5 + 2/6 x 4/5 = 2/3 x 2/5 + 1/3 x 4/5 = 4/15 + 4/15 = 8/15.

Answer: 8/15

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