Why does OG-11 PS Q-217 calculate prob. with only one possibility? That is, it assumes that the first student will be selected from the junior class with the porb. of 60/1000, and the prob. of selecting 2nd student will then be 1/800. Why not it takes into account the other possibility that first student is selected from senior class with the prob. of 60/800 and then the prob. of 2nd will be 1/1000. And then add the two possibilities to reach to a final prob. After all prob. is all about adding all the possibilities.
Pl. guide.
Nauman
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- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
- Pravin
Numan,
This problem is not about adding the individual probabilities it is about multiplying them. Read the question - it says both should be siblings -
Probability of selecting Junior class sibiling pair - 6/1000
Probability of selection Senior of the earlier selected sibling - 1/800
Total probability - 6/1000 * 1/800
Or do it like this
Probability of selecting Senior class sibiling pair - 6/800
Probability of selection of Junior the earlier selected sibling - 1/1000
Total probability - 6/800 * 1/1000
You get the same answer
This problem is not about adding the individual probabilities it is about multiplying them. Read the question - it says both should be siblings -
Probability of selecting Junior class sibiling pair - 6/1000
Probability of selection Senior of the earlier selected sibling - 1/800
Total probability - 6/1000 * 1/800
Or do it like this
Probability of selecting Senior class sibiling pair - 6/800
Probability of selection of Junior the earlier selected sibling - 1/1000
Total probability - 6/800 * 1/1000
You get the same answer
- JonathanSchneider
- ManhattanGMAT Staff
- Posts: 370
- Joined: Sun Oct 26, 2008 3:40 pm
Nauman, I imagine at this point you're asking: "right, but why don't we add those two together?"
The answer is that when we are using the "slot method" for finding probability, we are really listing a specific permutation, not just a combination. When listing a combination (such as two odds and one even), we would have to add the individual scenarios in order to find the number of permutations. But when we use the slot method we are picking IN ORDER; aka we are picking a permutation. As a result we don't need to go any further; we've already found the total number of specific options.
The answer is that when we are using the "slot method" for finding probability, we are really listing a specific permutation, not just a combination. When listing a combination (such as two odds and one even), we would have to add the individual scenarios in order to find the number of permutations. But when we use the slot method we are picking IN ORDER; aka we are picking a permutation. As a result we don't need to go any further; we've already found the total number of specific options.
5 posts Page 1 of 1