Problem Solving

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A.1/4
B.1/2
C.5/8
D.2/3
E.3/4

OA C

selango wrote:A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A.1/4
B.1/2
C.5/8
D.2/3
E.3/4

OA C

took me a while to get my head around this one.

the fixed probability is only on the fourth ball which should be black, no restriction on the rest: therefore: if the fourth must be black we start with only 7 balls to choose from and it is without replacement so 1 ball is reduced with every ball we take out.:

1st ball= 7/8, second ball= 6/7, third ball=5/6, fourth ball 5/5, fourth ball has a chance of being any of the 5 black balls out of 5 the remaining 5 balls.

7/8*6/7*5/6*5/5 [spoiler]which gives u 5/8, option C[/spoiler]

Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.

kvcpk wrote:Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.

Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!

This is so because each ball is equally likely to be the fourth one (or nth one) drawn

GMATGuruNY wrote:

kvcpk wrote:Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.

Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!

Thanks for sharing it Mitch... Very Useful

GMATGuruNY wrote:

kvcpk wrote:Surprisingly, Answer lloks to be same as picking the black ball in first attempt.

Is there any rule that the probability will be the same throughout? Sounds awkward.. but want to confirm.

Yes, P(B) on any given pick = P(B) on the first pick.

An easier example:

Say we have 3 red marbles and 2 blue. What's the probability that the second marble picked is blue?

P(B) on the first pick is 2/5.

To get blue on the second pick, our two picks could be RB or BB:

P(R and B) = 3/5 * 2/4 = 3/10.

P(B and B) = 2/5 * 1/4 = 1/10.

Since either is a good outcome, we add the fractions:

3/10 + 1/10 = 4/10 = 2/5.

The same probability as getting blue on the first pick.

Pretty neat!

Using this logic, why would the other problem be solved this way:
P(black first pick) 5/8
P (black 2nd pick) 4/7
P (black 3rd pick) 3/6
P (black 4th pick) 2/5
I think you'd have to add the first three and then multiply the black 4th pick ... but this gets really complicated. I guess I'm just trying to figure out how to do the two problems the same way, since we didnt use this methology to anwser the yellow /black ball problem.

singhpreet1 wrote:
took me a while to get my head around this one.

the fixed probability is only on the fourth ball which should be black, no restriction on the rest: therefore: if the fourth must be black we start with only 7 balls to choose from and it is without replacement so 1 ball is reduced with every ball we take out.:

1st ball= 7/8, second ball= 6/7, third ball=5/6, fourth ball 5/5, fourth ball has a chance of being any of the 5 black balls out of 5 the remaining 5 balls.

7/8*6/7*5/6*5/5 [spoiler]which gives u 5/8, option C[/spoiler]

why is the 1st pick 7/8. shouldnt it be 8/8 because anything is a favorable outcome since there is no restriction.

it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.

missrochelle wrote:it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.

Here are all the independent probabilities that would get us a good outcome (black on the 4th pick):

P(BBBB) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

P(YBBB) = 3/8 * 5/7 * 4/6 * 3/5 = 3/28
Since (BYBB) and (BBYB) are also good outcomes, total number of ways to get 1 yellow with black on the 4th pick = 3 * 3/28 = 9/28

P(YYBB) = 3/8 * 2/7 * 5/6 * 4/5 = 1/14
Since (BYYB) and (YBYB) are also good outcomes, total number of ways to get 2 yellow with black on the 4th pick = 3 * 1/14 = 3/14

P(YYYB) = 3/8 * 2/7 * 1/6 * 5/5 = 1/56

Since all of the above are good outcomes, we add the fractions:

1/14 + 9/28 + 3/14 + 1/56 = 35/56 = 5/8.

Much easier just to remember the rule I posted above:

P(B) on the nth pick is the same as P(B) on the 1st pick: 5/8.

GMATGuruNY wrote:

missrochelle wrote:it appears as though they are "reserving" the black ball for the fourth spot. therefore, it CANNOT be picked on the first pick, because it is "reserved" as the outcome of the 4th pick. hence, the use of "fixed probability" -- a new term for me!

however, im not sure how this method relates to the method posted by GMATGuruNY. One seems to be isolating the black ball event to 4th pick, while the other explanation goes through the independent probablities using AND/OR rules (Multiply/Add). I think the second method would be very arduous on this problem but I'm very curious to see how it would work out.

Here are all the independent probabilities that would get us a good outcome (black on the 4th pick):

P(BBBB) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

P(YBBB) = 3/8 * 5/7 * 4/6 * 3/5 = 3/28
Since (BYBB) and (BBYB) are also good outcomes, total number of ways to get 1 yellow with black on the 4th pick = 3 * 3/28 = 9/28

P(YYBB) = 3/8 * 2/7 * 5/6 * 4/5 = 1/14
Since (BYYB) and (YBYB) are also good outcomes, total number of ways to get 2 yellow with black on the 4th pick = 3 * 1/14 = 3/14

P(YYYB) = 3/8 * 2/7 * 1/6 * 5/5 = 1/56

Since all of the above are good outcomes, we add the fractions:

1/14 + 9/28 + 3/14 + 1/56 = 35/56 = 5/8.

Much easier just to remember the rule I posted above:

P(B) on the nth pick is the same as P(B) on the 1st pick: 5/8.

Very well explained!!!

P(B) on the nth pick is the same as P(B) on the 1st pick

Can you explain what this rule actually means in words and when is it applicable?

I followed an easier approach and arrived at the same answer.

First, fixing the 4th pick as a black, we have 7 balls (3 Y + 4 B) remaining. Using the ANAGRAMS technique,

P (B on the 4th) = all arrangements of YYYBBBB / all arrangements of YYYBBBBB

Numerator = 7!/3!4! = 35
Denominator = 8!/3!5! = 56

Answer = 35/56 = 5/8

nox104 wrote:

P(B) on the nth pick is the same as P(B) on the 1st pick

Can you explain what this rule actually means in words and when is it applicable?

I followed an easier approach and arrived at the same answer.

First, fixing the 4th pick as a black, we have 7 balls (3 Y + 4 B) remaining. Using the ANAGRAMS technique,

P (B on the 4th) = all arrangements of YYYBBBB / all arrangements of YYYBBBBB

Numerator = 7!/3!4! = 35
Denominator = 8!/3!5! = 56

Answer = 35/56 = 5/8

Here's the rule:

P(X) on the nth pick = P(X) on the 1st pick.

So if we have 3 yellow and 5 black marbles:

P(B) on the 1st pick is 5/8
P(B) on the 2nd pick is 5/8
P(B) on the 3rd pick is 5/8
P(B) on the 4th pick is 5/8

P(B) is the same for each pick.

Does this help?

Well I figured that it would be 5/8 each time from "P(B) on the nth pick is the same as P(B) on the 1st pick". But I was kinda trying to visualize it and understand why it is so..

Never mind, I am probably over-analyzing it. Thanks for your reply.

nox104 wrote:Well I figured that it would be 5/8 each time from "P(B) on the nth pick is the same as P(B) on the 1st pick". But I was kinda trying to visualize it and understand why it is so..

Never mind, I am probably over-analyzing it. Thanks for your reply.

Algebraically:

P(B) on the 1st pick = B/(B+Y)

P(B) on the 2nd pick:
P(BB) = B/(B+Y) * (B-1)/(B+Y-1) = (B^2-B)/[(B+Y)(B+Y-1)]
P(YB) = Y/(B+Y) * B/(B+Y-1) = BY[(B+Y)(B+Y-1)]
Adding, we get P(B) on 2nd pick:
= (B^2-B)/[(B+Y)(B+Y-1)] + BY[(B+Y)(B+Y-1)]
= (B^2-B+BY)/[(B+Y)(B+Y-1)]
=[B(B-1+Y)]/[(B+Y)(B+Y-1)]
=B/(B+Y)
=P(B) on the 1st pick

Hence, P(B) on the 2nd pick = P(B) on the 1st pick.

From this result we can predict that P(B) on the nth pick = B/(B+Y) = P(B) on the 1st pick.

Does this help?