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(edited 1 month ago)
Reply 1
Original post by username17263528
https://isaacphysics.org/questions/poisson_woodland?stage=a_level
Please could I have some help with part e. I tried to find the answer by finding the probability of it containing exactly 8 trees, the probability of it containing more than 8 trees and the probability of it containing less than 8 trees. I multiplied all of these together (including squaring the probability of it containing exactly 8), but when I submitted my answer, Isaac told me to consider the number of ways that this could be done.
I’m not sure how to account for this. Is this a case of conditional probability? Any help would be gratefully received. Thank you in advance.

Not really worked it through, but Id imagine is like the nCr type argument when you do binomial, so if youve worked out the probability that 2 out of 4 are 8, then how many combinations of 2 out of 4 are there and similarly for the other condition.
Original post by username17263528
https://isaacphysics.org/questions/poisson_woodland?stage=a_level
Please could I have some help with part e. I tried to find the answer by finding the probability of it containing exactly 8 trees, the probability of it containing more than 8 trees and the probability of it containing less than 8 trees. I multiplied all of these together (including squaring the probability of it containing exactly 8), but when I submitted my answer, Isaac told me to consider the number of ways that this could be done.
I’m not sure how to account for this. Is this a case of conditional probability? Any help would be gratefully received. Thank you in advance.

It's not conditional probability. If we write E for equals 8, L for less than 8 and M for more than 8, then

P(E)^2 P(L) P(M) would be the probability that the first 2 regions = 8, the 3rd region is < 8 and the 4th region is > 8 (or to put it another way, that EELM occurred in that order).

You need to account for all possible arrangements, e.g. EEML, ELME etc. (Hopefully it's clear that each arrangement will have the same probability, so at least you don't have to keep working that out for each arrangement).
Reply 3
Original post by username17263528
Thank you both. I’ve got the correct answer now. I had tried doing your method earlier but I’d made the mistake of saying that the total number of possible ways to do it was 8 instead of 12. Thanks again.

The "easiest" way of working 12 out was 4 for M, 3 for L and the two E must lie in the other 2, so 4*3.

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