Problem Solving

Compare these two questions:

1. Timmy has a bag of 5 candy bars, 3 are dark chocolate and 2 are milk chocolate. Assuming he picks two chocolate bars simultaneously and at random, what is the chance that exactly 1 of the bars he has picked is dark chocolate?

2. Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

1. For the first one, there are two possible scenarios: P(dark, milk) and P(milk, dark). Thus it would be 3/5 * 2/4 = 6/20. 2/5 * 3/4 = 6/20. So total probability would be 6/20 + 6/20 = 12/20 or 3/5.

2. The probability that any 1 letter would be put into the correct envelope and the other letters would be put into the wrong envelope is 1/4 * 2/3 * 1/2 * 1= 1/12. (After the first try, you would have three envelopes left and three letters. Since one of the letter goes into the second envelope, the probability of any of the remaining letters left will go into the wrong envelope is 2/3. This logic carries on for 1/2 and 1 as well). To get the probability for all four letters, it would be 4 * 1/12 = 1/3. Now this is the part where I am confused.

1/4 * 2/3 * 1/2 * 1 = 1/12. Or the probability for R-W-W-W (Let "R"= the probability that the right letter goes into the right envelope and "W"= the probability that the wrong letter goes into the wrong envelope.) So why aren't we calculating for the different orders of R-W-W-W? Such as,
W-R-W-W? (Like we would calculate for the different ways of getting only 1 dark chocolate bar on the first two tries?)

For example: W-R-W-W would be: 3/4 * 1/3 * 1/2 * 1 = 1/8
W-W-R-W would be: 3/4 * 2/3 * 1/2 * 1 = 1/4
W-W-W-R would be: 3/4 * 2/3 * 1/2 * 1 = 1/4.
So 1/12 +1/8 + 1/4 + 1/4= 17/24. Since there are four different letters, 4 * 17/24 = 68/24. This is definitely wrong since no probability can be greater than one. But what is wrong with my logic for the second problem? What is the differences between the first problem and the second problem?

Thanks! Your inputs are greatly appreciated!

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OneTwoThreeFour wrote:Compare these two questions:

1. Timmy has a bag of 5 candy bars, 3 are dark chocolate and 2 are milk chocolate. Assuming he picks two chocolate bars simultaneously and at random, what is the chance that exactly 1 of the bars he has picked is dark chocolate?

2. Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

1. For the first one, there are two possible scenarios: P(dark, milk) and P(milk, dark). Thus it would be 3/5 * 2/4 = 6/20. 2/5 * 3/4 = 6/20. So total probability would be 6/20 + 6/20 = 12/20 or 3/5.

2. The probability that any 1 letter would be put into the correct envelope and the other letters would be put into the wrong envelope is 1/4 * 2/3 * 1/2 * 1= 1/12. (After the first try, you would have three envelopes left and three letters. Since one of the letter goes into the second envelope, the probability of any of the remaining letters left will go into the wrong envelope is 2/3. This logic carries on for 1/2 and 1 as well). To get the probability for all four letters, it would be 4 * 1/12 = 1/3. Now this is the part where I am confused.

1/4 * 2/3 * 1/2 * 1 = 1/12. Or the probability for R-W-W-W (Let "R"= the probability that the right letter goes into the right envelope and "W"= the probability that the wrong letter goes into the wrong envelope.) So why aren't we calculating for the different orders of R-W-W-W? Such as,
W-R-W-W? (Like we would calculate for the different ways of getting only 1 dark chocolate bar on the first two tries?)

For example: W-R-W-W would be: 3/4 * 1/3 * 1/2 * 1 = 1/8
W-W-R-W would be: 3/4 * 2/3 * 1/2 * 1 = 1/4
W-W-W-R would be: 3/4 * 2/3 * 1/2 * 1 = 1/4.
So 1/12 +1/8 + 1/4 + 1/4= 17/24. Since there are four different letters, 4 * 17/24 = 68/24. This is definitely wrong since no probability can be greater than one. But what is wrong with my logic for the second problem? What is the differences between the first problem and the second problem?

Thanks! Your inputs are greatly appreciated!

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Intuitively, you probably feel that there should not be any difference between getting the first one right, or the second one right, or third, or fourth - the probabilities should be symmetrical.

One way to understand the problem with your proposed approach is that for W-R-W-W, the first W is not really a 3/4 probability: If you want to get the first one wrong and the second one right, then the first one actually has TWO "forbidden options": its own envelope, and the "right envelope", since we save that for the second place. Thus, there are only two wanted "wrong" envelopes, and the real probability of your case is
2/4*1/3*1/2*1 = 1/12.

The same for W-W-R-W : each W coming before the R must reduce one "wanted outcome" to save the right envelope for the R, so prob is 2/4*1/3*1/2*1 = still 1/12.

This is indeed unusual - when breaking down problems into one step at a time, we don't normally look "ahead" - only behind. The real difference between the cases is that in the second case you are not really counting order of choosing - you're finding the probability of a certain address being the right one. It's not REALLY 1st, 2nd, 3rd, 4th in terms of order of choosing, but rather p9address 1 being right, then others wrong, then p(address 2 being right, others wrong), etc. etc. The labels for your steps are not
1st, 2nd, 3rd, 4th

but rather

address A (right), address B (wrong), address C (wrong), address D (wrong).
1/4 * 2/3 * 1/2 * 1
and the right address can always be "swiveled" to be the "first" in the row - the order does not matter.

The second scenario would be
address B (right), address A (wrong), address C (wrong), address D (wrong).
1/4 * 2/3 * 1/2 * 1

Thank you Geva and Whit! Those are some very helpful advice! I think I was confused between "distinct" and "general" sequential probability. If I were to come up with another problem, such as

"Given we have four variables, a,b,c,d, what is the probability of choosing a & b on the first two tries"

Then it would be simply be: 2/4 * 1/3 = 1/6.

However in the problem with Timmy and chocolate bars, " Timmy has a bag of 5 candy bars, 3 are dark chocolate and 2 are milk chocolate. Assuming he picks two chocolate bars simultaneously and at random, what is the chance that exactly 1 of the bars he has picked is dark chocolate?"

Then I will have to find p(dark, milk) and p(milk, dark) and then add them together.

OneTwoThreeFour wrote:Thank you Geva and Whit! Those are some very helpful advice! I think I was confused between "distinct" and "general" sequential probability. If I were to come up with another problem, such as

"Given we have four variables, a,b,c,d, what is the probability of choosing a & b on the first two tries"

Then it would be simply be: 2/4 * 1/3 = 1/6.

However in the problem with Timmy and chocolate bars, " Timmy has a bag of 5 candy bars, 3 are dark chocolate and 2 are milk chocolate. Assuming he picks two chocolate bars simultaneously and at random, what is the chance that exactly 1 of the bars he has picked is dark chocolate?"

Then I will have to find p(dark, milk) and p(milk, dark) and then add them together.

Would just like to point out the two cases above are one and the same: the abcd problem can be set up as

a,b,don't care, don't care = 1/4*1/3*1*1
+
b,a,don't care, don't care = 1/4*1/3*1*1

total of

2*1/4*1/3 - which is the same ass your setup of 2/4*1/3.