Last visit was: 27 Apr 2024, 09:30 It is currently 27 Apr 2024, 09:30

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11691 [16]
Given Kudos: 136
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11691 [14]
Given Kudos: 136
Send PM
General Discussion
avatar
Intern
Intern
Joined: 05 Jun 2021
Posts: 9
Own Kudos [?]: 9 [3]
Given Kudos: 1
Send PM
Intern
Intern
Joined: 09 Jul 2018
Posts: 49
Own Kudos [?]: 77 [1]
Given Kudos: 0
Send PM
Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
1
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
Intern
Intern
Joined: 01 Feb 2022
Posts: 4
Own Kudos [?]: 1 [0]
Given Kudos: 11
Send PM
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?
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 11691 [1]
Given Kudos: 136
Send PM
Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
1
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
Intern
Intern
Joined: 10 Jun 2023
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 1
Send PM
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?
Verbal Expert
Joined: 18 Apr 2015
Posts: 28658
Own Kudos [?]: 33140 [0]
Given Kudos: 25178
Send PM
Re: Box A contains 2 black chips. Box B contains 2 white chips. [#permalink]
Expert Reply
it could pick the black form box C alike
Intern
Intern
Joined: 10 Jun 2023
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 1
Send PM
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!
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne