MAGOOSH Probability Problem

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pkw209: Master | Next Rank: 500 Posts; Posts: 266; Joined: Mon Oct 19, 2009 9:46 pm; Thanked: 8 times; GMAT Score:690

by pkw209 » Tue Feb 09, 2010 2:48 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?

The answer is 5/8.

Regardless of what ball was picked would the answer be 5/8 for black and 3/8 for yellow?

Can someone please explain?

Thanks.

Source: Beat The GMAT — Problem Solving | Tue Feb 09, 2010 2:48 pm

ldoolitt: Master | Next Rank: 500 Posts; Posts: 184; Joined: Sat Apr 14, 2007 9:23 am; Location: Madison, WI; Thanked: 17 times

by ldoolitt » Tue Feb 09, 2010 3:33 pm

pkw209 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?

The answer is 5/8.

Regardless of what ball was picked would the answer be 5/8 for black and 3/8 for yellow?

Can someone please explain?

Thanks.

You're correct. You could see this by applying the total probability theorm. For example on ball selection 2

P(Ball 2 is black) = P(Ball 2 is black GIVEN Ball 1 is black) * P(Ball 1 is black) + P(Ball 2 is black GIVEN Ball 1 is yellow) * P(Ball 1 is yellow)

Plugging in numbers

P(Ball 2 is black) = (4/7) * (5/8) + (3/8) * (5/7) = 5/8

This is true for any ball selection N given that you don't know what was picked on any turn prior to N.

pkw209: Master | Next Rank: 500 Posts; Posts: 266; Joined: Mon Oct 19, 2009 9:46 pm; Thanked: 8 times; GMAT Score:690

by pkw209 » Tue Feb 09, 2010 3:40 pm

Thank you. That is an excellent explanation.

Could you please provide the generic total probability theorem?

pkw209: Master | Next Rank: 500 Posts; Posts: 266; Joined: Mon Oct 19, 2009 9:46 pm; Thanked: 8 times; GMAT Score:690

by pkw209 » Tue Feb 09, 2010 3:43 pm

Just pulled this from the web. I think it looks right.

Given n mutually exclusive events A_1, ..., A_n whose probabilities sum to unity, then
P(B)=P(B|A_1) * P(A_1)+...+P(B|A_n) * P(A_n),

where B is an arbitrary event, and P(B|A_i) is the conditional probability of B assuming A_i.

ldoolitt: Master | Next Rank: 500 Posts; Posts: 184; Joined: Sat Apr 14, 2007 9:23 am; Location: Madison, WI; Thanked: 17 times

by ldoolitt » Tue Feb 09, 2010 4:23 pm

pkw209 wrote:Just pulled this from the web. I think it looks right.

Given n mutually exclusive events A_1, ..., A_n whose probabilities sum to unity, then
P(B)=P(B|A_1) * P(A_1)+...+P(B|A_n) * P(A_n),

where B is an arbitrary event, and P(B|A_i) is the conditional probability of B assuming A_i.

Yep thats it alright! You can also wikipedia it for reference purposes.

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Feb 09, 2010 7:14 pm

pkw209 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?

The answer is 5/8.

Regardless of what ball was picked would the answer be 5/8 for black and 3/8 for yellow?

Can someone please explain?

Thanks.

We can also reason this out.

If the answer was less (or more) than 5/8, that means that the chance of getting a black ball on the 4th pick would be different than getting a black ball on another pick. Without knowing the other results, is there any reason to believe that pick 4 is somehow special?

Here's another way to think about it:

If the answer was less (or more) than 5/8, then the chance of getting a yellow ball on pick 4 would be more (or less) than 3/8. Is there any reason why the probability of black would go up and yellow would go down (or vice-versa)?

The answer to both questions is "of course not", which is why, without knowing other results, the chance remains constant for each pick.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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sspms2002: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Tue Dec 29, 2009 4:36 am; Location: Pune

by sspms2002 » Tue Mar 30, 2010 12:56 pm

Great Explaination Stuart!!!

newton9: Senior | Next Rank: 100 Posts; Posts: 71; Joined: Wed Mar 17, 2010 1:31 pmFollowed by: 1 member; Thanked: 5 times

by newton9 » Tue Mar 30, 2010 1:46 pm

Hii Stuart,

I have a question here.

I agree with your explanation when the balls are replaced.

But here the balls are not replaced. Does it still hold good??

Say for example in first 3 attemps, I got black balls.

So there are 2 black balls and 3 Yellow balls left. How can the probability be 5/8 in this case? Is it not 2/5.

Thanks in advance.

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Mar 30, 2010 8:49 pm

newton9 wrote:Hii Stuart,

I have a question here.

I agree with your explanation when the balls are replaced.

But here the balls are not replaced. Does it still hold good??

Say for example in first 3 attemps, I got black balls.

So there are 2 black balls and 3 Yellow balls left. How can the probability be 5/8 in this case? Is it not 2/5.

Thanks in advance.

Hi,

replacement is irrelevant; the same logic applies.

If the answer is less than 5/8, that means that the chance of selecting a yellow ball 4th is greater than 3/8 (it has to be either black or yellow, right?); if the answer is greater than 5/8, that means that the chance of selecting a yellow ball 4th is less than 3/8. There's no logical reason why the probabilities would chance from the original chances.

It's only if we actually know the previous results that the probability would change. For example, if the question had been:

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. If the first 3 balls chosen are black, what's the probability that the fourth ball selected is black?

In the question at hand, we have to consider every possible case that includes the black ball as the #4 selection; when we add those up, we'll get 5/8 for black and 3/8 for yellow.

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