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# S2 poisson distribution question Watch

1. Two random variables X and Y have independent poisson distributions given by X~Poisson(1.4) and Y~Poisson(3.6) respectively.
Using the distributions of X and Y only, calculate P(X+Y =2).

I tried working it out doing this:
P(X+Y =2) = P(X=2) * P(Y=2)
0.241 * 0.177 = 0.042

However the answer in the book is 0.084 which is double what I got. Where have I gone wrong?
2. (Original post by IDontKnowReally)
Two random variables X and Y have independent poisson distributions given by X~Poisson(1.4) and Y~Poisson(3.6) respectively.
Using the distributions of X and Y only, calculate P(X+Y =2).

I tried working it out doing this:
P(X+Y =2) = P(X=2) * P(Y=2)
0.241 * 0.177 = 0.042

However the answer in the book is 0.084 which is double what I got. Where have I gone wrong?
But when x =2 and y= 2, x+y = 4.

Think about the different cases of x and y. As a starting point, it is immediate that x or y should each never be 3 or greater, if you can see why from the question.
3. Because they're independent, you can just add their means and then find that probability that X=2 of that mean.

[I'm also still learning this, so correct me if I'm wrong].
4. (Original post by Synmatic)
Because they're independent, you can just add their means and then find that probability that X=2 of that mean.

[I'm also still learning this, so correct me if I'm wrong].
Question does say "Using the distributions of X and Y only".

That aside, you are correct in that you can add their parameters.
5. (Original post by SeanFM)
But when x =2 and y= 2, x+y = 4.

Think about the different cases of x and y. As a starting point, it is immediate that x or y should each never be 3 or greater, if you can see why from the question.
How can I find what x and y are? Is it to do with the ratio of their means?
6. (Original post by IDontKnowReally)
How can I find what x and y are? Is it to do with the ratio of their means?
No, as above you can combine them using a property that is not required or used at A-level, and the question states that you cannot do this anyway. (I only mention this because you mentioned ratio, it's not quite ratio but the sum of the means eg if X~Po(10) and Y~Po(20) then X+Y~Po(30) but I repeat that this is not something you should know and the question says you are not allowed to use this method)

You tell me, what possible x and y values can you have in the equation x + y = 2?
7. (Original post by SeanFM)
No, as above you can combine them using a property that is not required or used at A_level, and the question states that you cannot do this anyway.

You tell me, what possible x and y values can you have in the equation x + y = 2?
Anything greater than 0 and anything less than 2?
8. (Original post by IDontKnowReally)
Anything greater than 0 and anything less than 2?
That is right, but you just have to be careful. It can also include 2.

Can you now see how to solve the question?
9. (Original post by SeanFM)
That is right, but you just have to be careful. It can also include 2.

Can you now see how to solve the question?
No, im still a little confused, sorry.
Would it just be P(X<=2) * P(Y<=2) ?
10. I would just add the 2 distributions so you would let Z = X + Y so then Z~Po(5)
Then use your method to find the P(Z=2), it should give the answer the book gives.

Also, your method could be right (I'm not sure) as there are 2 ways in which X and Y can happen, X first then Y or Y first then X. You only did it one of those ways which would explain your answer being half the answer in the book.
11. (Original post by IDontKnowReally)
No, im still a little confused, sorry.
Would it just be P(X<=2) * P(Y<=2) ?
No, because in that probability if I selected, for example, X=2 and Y = 2 (which are in each of those probabilities) then X+Y = 4, which does not satisfy X+Y = 2, so that is a hint for you

So what are all the x and y values greater than or equal to 0 such that x + y = 2? And why is this useful for the question?
12. (Original post by SeanFM)
No, because in that probability if I selected, for example, X=2 and Y = 2 (which are in each of those probabilities) then X+Y = 4, which does not satisfy X+Y = 2, so that is a hint for you

So what are all the x and y values greater than or equal to 0 such that x + y = 2? And why is this useful for the question?
x=1, y=1
x=2. y=0
x=0, y=2
Could you do P(X=1) * P(Y=1) + P(x=2)* P(Y=0) + P(X=0) * P(Y=2) ?
13. (Original post by IDontKnowReally)
x=1, y=1
x=2. y=0
x=0, y=2
Could you do P(X=1) * P(Y=1) + P(x=2)* P(Y=0) + P(X=0) * P(Y=2) ?
Well done with just very small nudges in the right direction you have worked it out for yourself.
14. (Original post by SeanFM)
Well done with just very small nudges in the right direction you have worked it out for yourself.

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