What are the odds of this happening?

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Can someone please tell me the percentage of probability that the following event occurring? Furthermore, please show your work and explain how you came to the conclusion that you did.

You have a bag full of 22 balls. They are identical in size, shape, color, and weight; the only difference is that 11 have the number 1 on them, and 11 have the number 2 on them.

You are tasked with blindly drawing eleven of the balls, one at a time, recording the number on each of the balls you draw. Furthermore, you do not put the balls back, after you've drawn them. Instead, after the first draw, there will be 21 balls in the sack, and after that, 20, and then 19, and so on, until there are eleven balls in the sack.

After you have drawn eleven of the balls, what are the odds that all eleven of the ones you drew are all the same number?

Remember, don't just give me the answer; explain how you got there.

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seems a little too much like homework to me.

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DStebbins, you're having difficulty "translating" a word problem into an algebraic equation. So, here's my method of "translating" word problems into algebra.

• The word "AND" translates into multiplication (×).
• The word "OR" translates into addition (+).
• The words "MUST BE" translate into a probability (that is, the probability based on the circumstances given to you, which you've described to us).

At this point, you need to re-write/re-think the word problem to conform with the words used above:

• You seek the probability of an outcome of all 1's OR all 2's.
• So, for example, to draw all 1's, the first ball MUST BE a 1, AND the second ball MUST BE a 1, AND the third ... etc

And that's pretty much all you need to know. I have not given you the answer, I have not given you all the work you must do, and I have not computed even one probability. You must compute individual probabilities and combine them appropriately according to their relationship with each other as I have shown above. I don't like saying that anything is "easy" with math, so you won't hear that from me. However, I will say that this kind of math is one of methodically translating and calculating. Hopefully at some point, the reasoning behind my method will become intuitively obvious.

PS — DStebbins, please ask a moderator to move this to the Homework Help sub-forum.

Edited by ewmon
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Just wanted to note that the first ball, required to be a 1 $(\frac{11}{22})$ OR(+) a 2 $(\frac{11}{22})$ would show as a 100% probability $(\frac{22}{22})$.

Sorry, but that's where I'd imagine myself tripping up...

Then just multiply that probability by all the following ones..

$\frac{22\times10!}{22!\div11!} = \frac{79833600}{28158588057600}$

or $\frac{1}{352716}$

Edited by Marqq
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Just wanted to note that the first ball, required to be a 1 $(\frac{11}{22})$ OR(+) a 2 $(\frac{11}{22})$ would show as a 100% probability $(\frac{22}{22})$.

Sorry, but that's where I'd imagine myself tripping up...

Then just multiply that probability by all the following ones.

Marqq. you're absolutely correct: The probability of drawing a 1 or a 2 as the first ball is 1.000 (ie, unity, or an absolute certainty). This is a common misinterpretation of these kinds of problems, and I'm glad you raised it. The problem does not ask for this probability, but instead, it asks for the probability that {all the 11 drawn balls are 1s} OR(+) {all the 11 drawn balls are 2s}. To quote myself more precisely:

You seek the probability of an outcome of {all 1's} OR {all 2's}.

So, this is what I meant by taking a word problem and restating it with a particular form of English (or any other language).

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Ewmon - I really like your explanation and the idea of the need to translate from a word problem to a maths problem carefully. Marqq's example of a possible trip up point is also easily voided if one translates his deliberately bad maths back into words; the first ball must be a 1 OR a 2, AND, the second ball must be a 1 OR a 2 ... is clearly different, even in words, to the first ball MUST BE a 1, AND the second ball MUST BE a 1... OR the first ball MUST BE a 2, AND....

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