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A box contains only red chips, white chips, and blue chips.

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A box contains only red chips, white chips, and blue chips. [#permalink]

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A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will be either white or blue?

(1) The probability that the chip will be blue is 1/5.
(2) The probability that the chip will be red is 1/3.
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A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will be either white or blue?

The probability that the chip will be either white or blue, equals to the probability that the chip will NOT be red, thus P(white or blue)=1-P(red).

(1) The probability that the chip will be blue is 1/5. Not sufficient.

(2) The probability that the chip will be red is 1/3. P(white or blue)=1-P(red)=1-1/3=2/3. Sufficient.

Answer: B.
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Re: A box contains only red chips, white chips, and blue chips. [#permalink]

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New post 14 Dec 2012, 05:35
Walkabout wrote:
A box contains only red chips, white chips, and blue chips. If a chip is randomly selected from the box, what is the probability that the chip will be either white or blue?

(1) The probability that the chip will be blue is 1/5.
(2) The probability that the chip will be red is 1/3.


Too simple :)

To find : probability of white or blue (which is equivalent to 1-probability of red)

S1 : there is no probability of white -- INSUFF
S2 : red - 1/3
Therefore p(W/B) = \(1-1/3 = 2/3\)
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Re: A box contains only red chips, white chips, and blue chips. [#permalink]

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New post 12 Apr 2017, 08:31
Hello from the GMAT Club BumpBot!

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Re: A box contains only red chips, white chips, and blue chips. [#permalink]

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New post 14 May 2017, 01:54
For those wondering, you could also find this answer another way. The way presented above is the fastest, but, just in case you are wondering, like I was, you could also do the following:

find the value of P(white) by doing:

1 - ( P(B) + P(R) ) = P(W)

Then do:

P(W or B) = P(W) + P(B) - P( W and B)

here P (W and B) is equal to zero, because it is not independent, it is mutually exclusive. (it is mutually exclusive because if you get one red you cannot get one white for example).

so it is simply:

P(W or B) = P(W) + P(B) - 0

Here is the calculation with numbers:

1- ( 1/5 + 1/3) = 7/15 = P(W)

now do: P(W or B) = P(W) + P(B) - 0

7/15 + 1/5 - 0 = 2/3

:)

hope it helps!
I was wondering about this other method when I did the problem.

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Re: A box contains only red chips, white chips, and blue chips. [#permalink]

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New post 25 Jul 2017, 06:46
concept :

For second .

P(R) + P (B) + P (W) = 1

P (W) + P(B) = 1 - P(R)

P(R) IS GIVEN.
so 2 is sufficient.

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Re: A box contains only red chips, white chips, and blue chips.   [#permalink] 25 Jul 2017, 06:46
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