Jason L.

asked • 07/01/16

Probability Combinatorics: Balls in a Bag

There's 14 balls in a bag. 3 of them have a prize, 11 of them do not. You get three chances to pick a ball out of the bag. After each pull, the ball you pulled does not go back into the bag, leaving one less ball to pull from. What are the odds of pulling exactly one prize ball?

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Tom K. answered • 07/01/16

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C(3,1)*C(11,2)/C(14,3)  (choose 1 of 3 prizes and 2 of 11 non-prizes out of a total of choosing 3 of 14) =
3 * 11*10/2!/(14*13*12/3!) =
3*3*11*10/(14*13*12) =
3*11*5/(7*13*4) =
165/364 =
.453297
Note: this is a hypergeometric distribution problem.  In Excel, use =HYPGEOMDIST(1,3,3,14)
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Mark M.

I do not understand the relevance of the 11C2.
The number of possible combinations is not relevant, IMHO.
Only three balls can be selected. Only three combinations are pertinent: Y N N, N Y N, OR N N Y.
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07/01/16

Tom K.

There are 11 balls that do not have prizes.  I suggest you read about the hypergeometric distribution.  The numerator is a product of the number of ways the prizes can be selected times the number of ways the non-prizes can be selected; the denominator is the number of ways all can be selected.
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07/02/16

Mark M.

Yes, 11 have not prizes and 3 have prizes. Yet this is not about distributions. It is about pricking 3 and only 3 balls (you might reread the problem) on the first three draws. Calculation is not made on the probability of any one ball appearing, but on the appearance of a particular combination during the first three draws.
Also hypergeometric distribution, IMHO, is a rather esoteric topic for the general population of this site.
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07/02/16

Tom K.

Hyper geometric is all about sampling without replacement.  I can prove this answer using conditional probabilities if you like.  Picking 1 prize means picking the prize first, second, or third.  P(picking prize first) = 3/14 * 11/13 * 10/12; p(picking prize second) = 11/14* 3/13 * 10/12; p(picking prize third) = 11/14 * 10/13 * 3/12;  note how these fractions are all the same and add to my answer.
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07/02/16

Jim S.

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Good answer, Tom...
jim
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07/02/16

Jason L.

That looks like exactly what I was looking for.  Thanks, Tom!
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07/02/16

Jim S. answered • 07/01/16

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Jason,
      If the the ball drawn was replaced this would a Bernoulli trail (independent trials and success probability is constant)
then you would use at the binomial distribution. Since the success probability is not constant as mentioned in the other answers, you would use the Hypergeometric distribution. Some examples can be found at http://www.statisticshowto.com/hypergeometric-distribution-examples/
 
Jim
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Mark M.

Yet the ball is not replaced.
The probability is constant. If exactly one is a prize, exactly two have to be non-prize. No variation!
Hypergeomtric distribution is a rather heavy handed method.
I ponder why this problem prompts so many conflicting - and erroneous - solutions.
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07/01/16

Joseph C. answered • 07/01/16

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As I see this situation,
on draw 1 there are 3 prizes among 14 balls : chances of drawing one prize is 3/14 = 0.214 ≅ 1 in 4.7
 
if no prize is drawn, then
on draw 2 there will be 3 prizes among 13 balls : chance of drawing one prize is 3/13 = 0.231 ≅ 1 in 4.3
 
if no prize is drawn, then
on draw 3 there will be 3 prizes among 12 balls : chance of drawing one prize is 3/12 = 0.25 = 1 in 4 
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Mark M.

Yet what are the odds of drawing exactly one?
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07/01/16

Mark M. answered • 07/01/16

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2184 (14C3) ways of picking three balls from a group of 14.
The ways of getting exactly one prize:
Yes No No
No Yes No
No No Yes
 
Probability of getting exactly one prize ball in a pick of three:
 
3/2184 or 1/728
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Jason L.

I do not think that is correct (I think your answer is the odds of all 3 prize balls?).  If there was just one pull, then the odds would be 3/14.  Since there are 3 pulls, there is 1 fewer ball in the bag on each pull, and we only need 1 prize, then the number is certainly greater than 3/14.  Do you know how to do it this way?
 
Thanks for trying to help.
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07/01/16

Mark M.

Your suggestion would allow for all three balls to be prize balls.
Note that you want the odds of getting "exactly" one prize ball. 
I did give the probability (successes/all) rather than odds (successes/failures) - 3/2181 or 1/727
 
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07/01/16

Jason L.

I actually did write "exactly one prize ball" in the original post.
 
To clarify, what I am looking to find out are the odds that I would pull exactly one prize ball from the bag if I had three chances to pull a ball (noting that if I didn't pull a prize ball, then there would be one fewer ball in the bag for the next pull).  
 
The answer I came up with was about .575, but I wasn't sure I did it right:
 
1 - ((COMBIN(11,3) + COMBIN(10,3) +COMBIN(9,3)) / (COMBIN(14,3) + COMBIN(13,3) +COMBIN(12,3)) )
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07/01/16

Mark M.

Your first sentence and the following paragraph and incongruous.
The sentence is addressed in my solution.
The paragraph ("to clarify") describes the probability of picking prize on the first draw OR the second draw OR the third draw. These probabilities work only if the events are mutually exclusive. They are not. What you pick on the first draw affects what you get on the second.
Say on the first pick you get a non-prize, the probability for the second is 3/13. Yet this second draw could also be a non-prize. Then the probability of a prize on the third draw is 3/12. This only give the probability of getting a prize on a particular draw, not one out of a draw of three.
 
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07/01/16

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