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 LIKELYIn case i, the other chip is also black
In case ii, the other chip is also black
In case v, the other chip is whiteSo, 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 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.