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Probability

by knight247 » Tue Oct 04, 2011 11:47 pm
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?
1/4
1/2
1/2
5/8
2/3
3/4

Don't have an OA. Detailed explanations would be appreciated. Thanks
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by Anurag@Gurome » Wed Oct 05, 2011 12:25 am
knight247 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?
1/4
1/2
1/2
5/8
2/3
3/4

Don't have an OA. Detailed explanations would be appreciated. Thanks
YBBB; probability that 4th ball is black = 3/5 (as we have already picked 2 black balls, so the remaining are 3 balls)
YYBB; probability that 4th ball is black = 4/5 (as we have already picked 1 black ball, so the remaining are 4 balls)
YYYB; probability that 4th ball is black = 5/5 (as we have not picked any black ball, so there are 5 balls)
BBBB; probability that 4th ball is black = 2/5 (as we have already picked 3 black balls, so the remaining are 2 balls)

Therefore required probability = [(3C1 * 5C2)/8C3] * 3/5 + [(3C2 * 5C1)/8C3] * 2/5 + [3C3/8C3] * 1 + (5C3/8C3) * 2/5 = (30/56) * (3/5) + (15/56) * (4/5) + (1/56) + (10/56) * (2/5) = (18 + 12 + 1 + 4)/56 = 35/56 = [spoiler]5/8[/spoiler]
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by shankar.ashwin » Wed Oct 05, 2011 12:28 am
Also to add,

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

Hence 5/8
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by knight247 » Wed Oct 05, 2011 12:38 am
@Ashwin

Can u explain how that is so?
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by shankar.ashwin » Wed Oct 05, 2011 12:54 am
The method Anurag did it is how its usually done. You could do it that way and check for any sum.

But its intuitive to know, that say we pick up the first ball,

P(Y) is 3/8 and P(B) is 5/8.

Now say you want the second ball to be black,
The way you usually do it would be to consider 2 scenarios either a black or yellow was picked up in the first attempt.

So that would be (5/8)(4/7) + (3/8)(4/7) = 5/8 (First was 2 blacks and 2nd was yellow and then black).

The same pattern is repeated till the 8th ball, the individual prob of picking up the balls remains the same. One other way to think about it is, we dont know which balls were removed before you pick up the black ball the 4th time, each ball you remove could be either yellow or black, and again P(Y) is 3/8 and P(B) is 5/8. I guess you understand the pattern
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by Brent@GMATPrepNow » Wed Oct 05, 2011 5:44 am
shankar.ashwin wrote:Also to add,
P(B) on any given pick = P(B) on the first pick.
Hence 5/8
Exactly. Nice work, shankar.ashwin!

This reminds me of when I was young, and my friends and I would sometimes "draw straws" to randomly select one person for a task (usually a less than ideal task like getting wood for the fire, or something stupid like jumping off a roof).

So, someone would hold up n pieces of grass (for n guys), and one of those pieces of grass was very short. The person who selected the shortest piece was the one to do the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece was minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that all n guys have a 1/n chance of selecting the shortest piece, regardless of when they select.

The same applies to the original question here.

Cheers,
Brent
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by GMATGuruNY » Wed Oct 05, 2011 6:07 am
knight247 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?
1/4
1/2
1/2
5/8
2/3
3/4

Don't have an OA. Detailed explanations would be appreciated. Thanks
I posted a solution -- and showed why the correct answer can be determined with almost no work -- here:

https://www.beatthegmat.com/probablity-ques-t60161.html

Other problems testing the same concept:

https://www.beatthegmat.com/manhattan-pr ... 89481.html

https://www.beatthegmat.com/probability-t60530.html

https://www.beatthegmat.com/probabilty-q ... 68798.html
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As a tutor, I don't simply teach you how I would approach problems.
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