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# Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except

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Math Expert
Joined: 02 Sep 2009
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Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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29 Mar 2018, 22:40
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:49) correct 21% (02:38) wrong based on 88 sessions

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Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except for color. Bag 2 contains 8 blue, 2 gold, and 2 red chips, also all identical except for color. Two chips, one from each bag, are randomly drawn, one after the other. What is the probability that neither chip will be gold?

(A) 5/18
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

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Re: Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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29 Mar 2018, 22:59
Since we have to all 10 blue balls are identical to number of ways of fetching them =1
similarly for gold=1 and red=1
Since three different colours are there so picking one ball will be 3C1
from two different bag total numbers=3C1*3C1
except gold=2C1*2C1
probability=4/9
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Re: Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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29 Mar 2018, 23:16
Bunuel wrote:
Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except for color. Bag 2 contains 8 blue, 2 gold, and 2 red chips, also all identical except for color. Two chips, one from each bag, are randomly drawn, one after the other. What is the probability that neither chip will be gold?

(A) 5/18
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

10+8+6 = 24 no gold = 16/24
8+2+2=12 no gold = 10/12

no gold = 16/24 x 10/12 = 2/3x5/6 = 5/9
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Re: Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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29 Mar 2018, 23:57
Dear kunal
You have ignored the importance of the number of balls in the probability.
The total number of balks in first bag =24. Number of balls other than gold =16.
Probability if selecting a ball other than gold =16/24= 2/3
Similarly for second bag, total balls =12, balls other than gold = 10.
Hence probability of selecting balls other than gold = 10/12 =5/6.

Required probability = 2/3 *5/6 =5/9

kunalcvrce wrote:
Since we have to all 10 blue balls are identical to number of ways of fetching them =1
similarly for gold=1 and red=1
Since three different colours are there so picking one ball will be 3C1
from two different bag total numbers=3C1*3C1
except gold=2C1*2C1
probability=4/9

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Re: Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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30 Mar 2018, 00:15
Hi gmatbuster,

Since all same colour balls are identical total number dosent matter....
because even you pick any ball the ball will be same only so total number of ways will be 1.
24C1 will come into picture when we can differentiate the ball. So that there will 24+1(if ball is not picked)ways to pick a ball
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Re: Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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04 Apr 2018, 16:49
Bunuel wrote:
Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except for color. Bag 2 contains 8 blue, 2 gold, and 2 red chips, also all identical except for color. Two chips, one from each bag, are randomly drawn, one after the other. What is the probability that neither chip will be gold?

(A) 5/18
(B) 4/9
(C) 1/2
(D) 5/9
(E) 2/3

The probability that the chip from bag 1 is not a gold chip is 16/24 = 2/3.

The probability that the chip from bag 2 is not a gold chip is 10/12 = 5/6.

Since we want both events to occur (i.e., bag 1 not gold and bag 2 not gold), we multiply the two respective probabilities. Thus, the probability that neither chip will be gold is 2/3 x 5/6 = 10/18 = 5/9.

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Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except  [#permalink]

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18 Apr 2018, 00:36
I am going through this problem and wonder why can't I solve this problem as per below and still get the answer 5/9?
1 - (8/24x 2/12) probability that the chip from bag A is gold x probability that the chip from bag B is gold
Answer i get is 1 - 1/18 = 17/18
Bag 1 contains 10 blue, 8 gold, and 6 red chips, all identical except   [#permalink] 18 Apr 2018, 00:36
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