aritrar4 wrote:
Could you please help me figure out the mistake I'm making in my approach below?
We need probability of opening at least 1 box => Prob(open 1 box) + Prob(open 2 boxes)
P(1box) = 5C1 * 10/10 * 1/9 * 8/8 * 6/7
That is,
5C1 = ways to select a box to open
10/10 = Prob of Selecting 1st key (can select any one of 10 keys)
1/9 = Prob of Selecting 2nd Key (must select the pair key of the first selection to open a box)
8/8 = Prob of Selecting 3rd key (can select any one of 8 keys)
6/7 = Prob of Selecting 4th key (can select any one except the pair of the second key)
So, P(1Box) = 5C1 * 10/10 * 1/9 * 8/8 * 6/7 = 30/63
If you begin by picking a specific box that we'll open, so if you begin by multiplying by 5C1, then the probability the first key will open that box is 2/10. It's not true that we can just pick any key at all. So you're multiplying by an extra 5 that you shouldn't be multiplying by when your calculation begins with "5C1 * 10/10". But then you're also assuming the very first two keys we pick will match, and that the third and fourth keys will not. We could pick our matching keys in a few ways - maybe the second and fourth keys match, for example, and the first and third do not. It turns out there are 6 different ways in which we could pick the two matching keys, which is why, when you multiplied by the extra 5, you got an answer fairly close to the right one (if you multiplied by 6 instead, it would be perfect). There's a similar issue with the calculation you've done in the second case -- multiplying by 5C2 is not correct.
As I said above, I'd never consider solving this directly, because it's both difficult and error-prone. But this is how you can do it:
- first you can ask, "what is the probability the first two keys open a single box, and the next two keys we pick do not open the same box?" The first key can be anything, the second must be a specific key (1/9 probability), the third key can be anything, and the fourth must not match the third (6/7 probability). So the probability we get this exact sequence is (1/9)(6/7).
- but we might be able to open exactly one box in several ways. Maybe the first two keys match, or maybe the second and the fourth match, for example. There are 4C2 = 6 ways to choose which two of the four keys match, so the probability that we can open precisely one box is (6)(1/9)(6/7)
- then we need to add the probability we can open both boxes. Again, we can imagine the first two keys match, and the last two match. The probability that will happen is (1/9)(1/7), since the first and third keys can be anything.
- But there are other sequences where we can open both boxes - maybe the 1st and 2nd keys match, or maybe the 1st and 3rd, or maybe the 1st and 4th. So there are 3 possible sequences (the first key must match something, so these are all the cases). So the probability we can open both boxes is (3)(1/9)(1/7)
Adding the results from our two non-overlapping cases, the answer is
(6)(1/9)(6/7) + (3)(1/9)(1/7) = 12/21 + 1/21 = 13/21
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