Box A contains 2 black chips. Box B contains 2 white chips. Box C contains 1 black

[BTGModeratorVI]

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

Answer: E
Source: GMAT prep now

[Brent@GMATPrepNow]

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

Answer: E
Source: GMAT prep now

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

[Scott@TargetTestPrep]

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

Answer: E

Solution:

Let’s look at the possible outcomes and their respective probabilities. Let’s let A, B, or C stand for the box that is selected, and let’s let X = the outcome that a black chip is drawn. We see that the probability of selecting a particular box is ⅓ and that the probability of drawing a black chip from box A is 1, from box B is 0, and from box C is ½.

For box A, the outcome (A, X) has a probability of 1/3 x 1 = 1/3.

For box B, the outcome (B, X) has a probability of 1/3 x 0 = 0 (i.e., we can’t draw a black chip).

For box C, the outcome (C, X) has a probability of 1/3 x ½ = 1/6.

Now, the only way that the other chip in the box is also black requires that we selected box A. Thus, we have the conditional probability statement “Given that a black chip was drawn, what is the probability that we chose box A?”.

The total probability that a black chip was drawn is 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2. The probability that the box selected was box A will thus be 1/3 / 1/2 = 2/3.

Alternate Solution:

Since we know the selected chip is black, it is impossible for this chip to be drawn out of box B. Of the remaining two boxes, box A contains a greater number of black chips, therefore the probability that box A was selected should be greater than the probability that box C was selected. Since 2/3 is the only answer choice greater than 1/2, it must be the correct choice.

Answer: E