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Box A contains 2 black chips. Box B contains 2 white chips. : Multiple-choice Questions — Select One Answer Choice

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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
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GreenlightTestPrep wrote:
Box A contains 2 black chips. Box B contains 2 white chips. Box C contains 1 black chip and 1 white chip. Ted chooses a box at random and then randomly selects a chip from that box. If the selected chip is black, what is the probability that the other chip in the same box is also black?

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


Since the first chip selected is black, Box B is rendered irrelevant: the first chip must be selected from A or C.
Between Boxes A and C, there are 3 black chips and 1 white chip:
BBBW
Once a black chip has been selected from A and C, the following chips remain:
BBW
Since there are now 2 black chips and 1 white chip -- for a total of 3 chips -- the probability that the next chip selected is black = 2/3.

Show: ::
E
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
guestuser wrote:
GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
Box A contains 2 black chips. Box B contains 2 white chips. Box C contains 1 black chip and 1 white chip. Ted chooses a box at random and then randomly selects a chip from that box. If the selected chip is black, what is the probability that the other chip in the same box is also black?

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


Let’s let:
B1 be one black chip in Box A.
B2 be the other black chip in Box A.
W1 be one white chip in Box B.
W2 be the other white chip in Box B.
B3 be the black chip in Box C
W3 be the white chip in Box C

There are 6 EQUALLY LIKELY outcomes:
Case i: Ted selects B1 from box A
Case ii: Ted selects B2 from box A
Case iii: Ted selects W1 from box B
Case iv: Ted selects W2 from box B
Case v: Ted selects B3 from box C
Case vi: Ted selects W3 from box C

If the selected chip is black, then we’re dealing with cases i, ii, or v
All 3 cases are EQUALLY LIKELY
In case i, the other chip is also black
In case ii, the other chip is also black
In case v, the other chip is white


So, of the 3 possible cases, 2 cases are such that the other chip in the box is black.

P(other chip is black) = 2/3

Answer: E

Cheers,
Brent


I think the problem is a Bayesian probability.
P(event) is the probability of an event. And P(event 1 | event 2) is the conditional probability of event 1 on the occurrence of event 2.

P(the other chip in the same box is also black | the selected chip is black)
= P(the selected chip is black & the other chip in the same box is also black) / P(the selected chip is black).

P(the selected chip is black & the other chip in the same box is also black) = 1/3 <= case i, ii
P(the selected chip is black) = 2/3 as calculated in the quote.

Hence, D) 1/2.


I agree with your answer. Is it proven that the correct answer is D?
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
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guestuser wrote:

I think the problem is a Bayesian probability.
P(event) is the probability of an event. And P(event 1 | event 2) is the conditional probability of event 1 on the occurrence of event 2.

P(the other chip in the same box is also black | the selected chip is black)
= P(the selected chip is black & the other chip in the same box is also black) / P(the selected chip is black).

P(the selected chip is black & the other chip in the same box is also black) = 1/3 <= case i, ii
P(the selected chip is black) = 2/3 as calculated in the quote.

Hence, D) 1/2.


I've highlighted the error in your solution.

P(the selected chip is black) cannot equal 2/3. Here's why:
We have 1 all-white box, 1 all-black box, and 1 50-50 box.
Since white balls and black balls are equally represented, it must be the case that: P(we select a white ball) = P(we select a black ball)
Since P(we select a white ball) + P(we select a black ball) = 1, it must be the case that P(we select a white ball) = P(we select a black ball) = 1/2
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
GreenlightTestPrep wrote:
GreenlightTestPrep wrote:
Box A contains 2 black chips. Box B contains 2 white chips. Box C contains 1 black chip and 1 white chip. Ted chooses a box at random and then randomly selects a chip from that box. If the selected chip is black, what is the probability that the other chip in the same box is also black?

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


Let’s let:
B1 be one black chip in Box A.
B2 be the other black chip in Box A.
W1 be one white chip in Box B.
W2 be the other white chip in Box B.
B3 be the black chip in Box C
W3 be the white chip in Box C

There are 6 EQUALLY LIKELY outcomes:
Case i: Ted selects B1 from box A
Case ii: Ted selects B2 from box A
Case iii: Ted selects W1 from box B
Case iv: Ted selects W2 from box B
Case v: Ted selects B3 from box C
Case vi: Ted selects W3 from box C

If the selected chip is black, then we’re dealing with cases i, ii, or v
All 3 cases are EQUALLY LIKELY
In case i, the other chip is also black
In case ii, the other chip is also black
In case v, the other chip is white


So, of the 3 possible cases, 2 cases are such that the other chip in the box is black.

P(other chip is black) = 2/3

Answer: E

Cheers,
Brent


Someone clarify my doubt below:
The case in question is the probability of selecting one black chip on first pick and second black chip on the second pick.
Since the question mentions 'same box', there is only 1 box where this is possible i.e., box A where both the chips are black.
Rest all other cases don't qualify towards the probability.
Therefore that evaluates as 1 out 3 boxes. Hence probability is 1/3. What's wrong with my logic?
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
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it could pick the black form box C alike
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
Carcass wrote:
it could pick the black form box C alike

got it, thanks!
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
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The clue is in the question. Remember the AND/OR rule. Multiply probabilities when you see ‘and’. Add probabilities when you see ‘or’.

Think about the process. Draw a picture if it helps!

Let’s break it down. Ted has essentially 3 options.

Option 1 is that he chooses the first box (call it Box A) and the black chip inside.
Option 2 is that he chooses the second box (Box B) and the black chip inside.
Option 3 is that he chooses the third box (Box C) and the black chip inside.

Those options can be broken down using the AND/OR rule. Ted can choose either Box A, or Box B, or Box C.

So, the schema looks like:

(Box A * black chip) + (Box B * black chip) + (Box C * black chip)

The probability of choosing (Box A), (Box B) or (Box C) is (⅓) respectively.

The probability of choosing a black chip in box A is 100% (there is only one black chip in Box A).
The probability of choosing a black chip in box B is 0% (there are no black chips in Box B).
The probability of choosing a black chip in box C is 50% (half of the chips in Box C are black).

Therefore, plugging in the values into the formula gives you:

(⅓ * 1) + (⅓ * 0) + (⅓ * ½) = (⅓) + (⅙) =(½)

This is a relatively straightforward way to think about the question. If you are familiar with the AND/OR rule in probability, and how the rule applies, you can easily break this problem down and solve it within 60 seconds.

By calculating the eventualities iteratively, you could be stuck on this question for 90-120 seconds or even longer.

By the way… this is a layman's explanation of Bayes’ theorem and the probability formula. You will certainly save time on questions like this if you have a solid understanding of that formula and how to use it.

The Ultimate GRE Cheat Sheet has a comprehensive chapter on probability rules and time saving techniques. You must know these to get a top score on the GRE Quant. Go to Ultimate Tuition to download your copy.
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Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
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