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# Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he select

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Joined: 13 Apr 2013
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Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he select  [#permalink]

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16 Jul 2018, 10:01
1
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:19) correct 21% (01:59) wrong based on 19 sessions

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Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he selects three chips from the bag at random, what is the probability that all three chips selected are blue?

1. 1/28
2. 27/512
3. 9/64
4. 2/7
5. 167/168

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Re: Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he select  [#permalink]

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16 Jul 2018, 10:10
AkshdeepS wrote:
Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he selects three chips from the bag at random, what is the probability that all three chips selected are blue?

1. 1/28
2. 27/512
3. 9/64
4. 2/7
5. 167/168

Ways to pick up blue =6C3=5*4=20
Total ways to pick any three = 16C3=16*15*14/6=560

Prob = 20/560=1/28

A
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Re: Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he select  [#permalink]

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16 Jul 2018, 10:15
1
AkshdeepS wrote:
Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he selects three chips from the bag at random, what is the probability that all three chips selected are blue?

1. 1/28
2. 27/512
3. 9/64
4. 2/7
5. 167/168

Approach 1: Required Probability is of B B B = (6/16)*(5/15)*(4/14) = 1/28

Approach 2: 3 chips can be selected out of 16 chips in 16C3 ways

3 blue chips can be selected out of 6 blue chips in 6C3 ways.

Hence probability of selecting 3 blue chips out of 16 chips = 6C3/16C3 = $$\frac{(6*5*4)(1*2*3)}{(1*2*3)(16*15*14)}$$ = $$\frac{1}{28}$$

Thanks,
GyM
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Re: Bert has a bag of 6 Red, 6 Blue, and 4 White poker chips. If he select &nbs [#permalink] 16 Jul 2018, 10:15
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