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A little fuzzy on this probability concept

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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A little fuzzy on this probability concept

by bdevas01 » Thu Oct 14, 2010 9:22 am
There was a post on this site that discussed this topic but I couldn't find it so I'm posting it again.

A bag has 3 oranges and 2 apples, what is the probability of picking out an apple on the third try. I think the answer is 2/5, which is exactly the same probability as picking it out on the first try.

Why is this the case?
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by Brian@VeritasPrep » Thu Oct 14, 2010 10:52 am
Great question - and I love that you're asking "why is that the case" instead of just accepting it as a fluke anomaly or something like that...it's in that asking "why" that you tend to really build the kind of flexible knowledge that is such a benefit on this test.

If you wanted to calculate the sequences that would get you to a "third pick = apple" outcome, you can do that:

A, B, A = 2/5 * 3/4 * 1/3 = 1/10
B, B, A = 3/5 * 2/4 * 2/3 = 1/5
B, A, A = 3/5 * 2/4 * 1/3 = 1/10

Sum those up and you get 2/5, which like you said is the same as the initial probability. Here's at least one angle on why:

The "third draw" should have the same probability as the "third position" in a line, right? Whether you're drawing three pieces of fruit without replacement or just lining them up the probabilities shouldn't change. For example, if you were arranging the fruit on a table like:

Apple, Banana, Apple, Banana, Banana

or

Banana, Banana, Apple, Apple, Banana

The probability of having an apple in the third spot should be the same as the probability of drawing an apple third if you were picking them out of a bag, right? And in the lined-up-on-a-table case, it should stand to reason that the probability of an apple in any one position is 2/5 - no one position should have a different probability just because it's in a different spot.

So that's at least one demonstration of why the probability would be the same. The third "position" shouldn't have a different probability than the third "draw", and since that's the case it may be easier to calculate this one using the position model.
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by fskilnik@GMATH » Thu Oct 14, 2010 11:04 am
bdevas01 wrote: A bag has 3 oranges and 2 apples, what is the probability of picking out an apple on the third try?
Hello, bdevas01. Nice problem!

Let me have a shot!

First realize that you cannot have the first two selections as apples, therefore two possibilities may occur:

(a) You got two oranges at first two selections.
(b) You got one orange and one apple at first two selections.

Case (a) occurs with probability : 3/5 * 2/4 = 3/10 (easy) and in occuring this we still have 1 orange and 2 apples to choose for the third selection, meaning that the probability from now on is (case (a) assumed) is 2/3, and that means that the partial answer is therefore 3/10 * 2/3 = 1/5.

Case (b) occurs with probability : 2 * (3/5 * 2/4) = 3/5 (not "that" easy, think about it) and in occuring this we still have 2 oranges and 1 apple to choose for the third selection, meaning that the probability from now on is (case (b) assumed) is 1/3, and that means that the partial answer is therefore 3/5 * 1/3 = 1/5 again.

From the fact that Cases (in reality "events") (a) and (b) are mutually exclusive (one cannot occur when the other does), you can add the probabilities to get the 2/5 you expected.

There are more "elegant" ways to deal with it but this approach is certainly "GMAT" allowed (time restriction and elementary method being considered)...

Regards,
Fábio.
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by fskilnik@GMATH » Thu Oct 14, 2010 11:10 am
Brian@VeritasPrep wrote:And in the lined-up-on-a-table case, it should stand to reason that the probability of an apple in any one position is 2/5 - no one position should have a different probability just because it's in a different spot.
I thought about explaining why 2/5 was really the answer, but I missed the point of bdevas01´s question, for sure.

I guess what I put in quote (from Brian´s solution) is enough for the question asked!

Great insight, Brian, by the way.
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by bdevas01 » Thu Oct 14, 2010 3:55 pm
Brian and fskilnik, you guys hit it on the nail! Thank you so much for the prompt and detailed response!

I think the key take away here is that both the apples and the oranges have an equal chance of appearing in any of the five available slots and this is the reason why there is a 2/5 chance for the apple to be pulled on the third attempt and a 3/5 chance for the orange to be pulled from the bag. Also, note that that adding these two probabilities equals 1, which means that choosing an apple or an orange from the bad is CERTAIN on the third attempt.

Thanks again guys![/list]
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