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09 May 2014, 11:28
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Hey everyone, so I was wondering when you have to account for "order" in a probability question. For example:
Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket? [Veritas Prep]
A: 1/96
B: 1/48
C: 1/24
D: 1/12
E: 17/24
On this question I would think the answer would be 1/8 since you take (1/4)*(1/8)*(2/3)*3! given that the order in which you play isn't specified.
Here's another example where the order does matter:
Mathematics, physics, and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many mathematics books as physics books and the number of physics books is 4 greater than that of the chemistry books. Ricardo selects 1 book at random from the shelf, reads it in the library, and then returns it to the shelf. Then he again chooses 1 book at random from the shelf and checks it out in order to read at home. What is the probability Ricardo reads 1 book on mathematics and 1 on chemistry? [Veritas Prep]
A: 3%
B: 6%
C: 12%
D: 20%
E: 24%
So the answer ends up being 12% because you have a 6% chance * 2 ways to do this. Why isn't it the same for the question above? I'd love some input on this - thanks!
Penny, Leonard, and Sheldon each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/4, 1/8, and 1/3, respectively, what is the probability that Penny and Leonard, but not Sheldon, will make the basket? [Veritas Prep]
A: 1/96
B: 1/48
C: 1/24
D: 1/12
E: 17/24
On this question I would think the answer would be 1/8 since you take (1/4)*(1/8)*(2/3)*3! given that the order in which you play isn't specified.
Here's another example where the order does matter:
Mathematics, physics, and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many mathematics books as physics books and the number of physics books is 4 greater than that of the chemistry books. Ricardo selects 1 book at random from the shelf, reads it in the library, and then returns it to the shelf. Then he again chooses 1 book at random from the shelf and checks it out in order to read at home. What is the probability Ricardo reads 1 book on mathematics and 1 on chemistry? [Veritas Prep]
A: 3%
B: 6%
C: 12%
D: 20%
E: 24%
So the answer ends up being 12% because you have a 6% chance * 2 ways to do this. Why isn't it the same for the question above? I'd love some input on this - thanks!
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09 May 2014, 12:23
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For the first question, think of it this way. Suppose there was an order: P, L, then S.
Then the probability is (1/4)*(1/8)*(2/3), right?
Now let's change the order to S, L, P. Now the new probability is (2/3)*(1/8)*(1/4).
I guess what I'm trying to say is that the order doesn't matter, the probability will always be 1/48. There's really no reason to multiply these three probabilities by 3!. When you're picking things however, you would need to multiply by a 3!.
Then the probability is (1/4)*(1/8)*(2/3), right?
Now let's change the order to S, L, P. Now the new probability is (2/3)*(1/8)*(1/4).
I guess what I'm trying to say is that the order doesn't matter, the probability will always be 1/48. There's really no reason to multiply these three probabilities by 3!. When you're picking things however, you would need to multiply by a 3!.
20 May 2014, 06:18
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bekerman
Lets take the first case above and consider ordering as well:
As you very well no, there are 3! = 6 cases in which Penny (P), Leonard (L) and sheldon(S) can throw the basketball.
Now, probability of the order P->L->S is (1/4*1/8*2/3) *1/6 ---- you have to multiply by 1/6 because 6 total cases are possible and each case will have a chance of 1/6
Similarly rest of the 5 cases will have the same probability. Summing them up it will result in - (1/4*1/8*2/3) = 1/48
After some practice you will realize that you don't need to consider the ordering at all and just multiply the chance for each participant.
If the above question had an additional constrain that Penny should go first. We will have to consider the ordering and cannot do it directly like above.
The answer in that case will be 1/48*(2/6) = 1/144.....as penny will go first in 2 out of 6 cases
In question 2 on solving you will get: chemistry = 2, phy = 6 and Maths = 12
Now either Ricardo can read chemistry in library and take maths home OR read maths in library and take chemistry home. So probability = 2 *(2/20)*(12/20) = 12%
