[enigmaaa]

If X~Poisson (10),

a) use the poission table to find P() .

b) two successive values of X are now chosen independently from the distribution. Find the probability that just one of them is at least 7.

for a) I used 1- P() = 1 - 0.1301 = 0.8699. But answer in textbook said 0.7798, I think they used 1 - P() just to double check if I was right?

b)
Let A = greater than or equal to 7
B = less than 7

I considered 4 cases:
1.first value is A, second also A;
2.first value is A, second B;
3.first value is B, second A;
4.first value is B, second also B;

So it would be P(2)+P(3) = 0.8699 x 0.1301 + 0.8699 x 0.1301 = 0.226

But answer is 0.343, is it because they used the wrong value for the first time as well?

Many thanks.

[magdaplaysbass]

(Original post by enigmaaa)
If X~Poisson (10),

a) use the poission table to find P() .

b) two successive values of X are now chosen independently from the distribution. Find the probability that just one of them is at least 7.

for a) I used 1- P() = 1 - 0.1301 = 0.8699. But answer in textbook said 0.7798, I think they used 1 - P() just to double check if I was right?

b)
Let A = greater than or equal to 7
B = less than 7

I considered 4 cases:
1.first value is A, second also A;
2.first value is A, second B;
3.first value is B, second A;
4.first value is B, second also B;

So it would be P(2)+P(3) = 0.8699 x 0.1301 + 0.8699 x 0.1301 = 0.226

But answer is 0.343, is it because they used the wrong value for the first time as well?

Many thanks.

I get the same answers as you do
It must just be an error in your book. Like you said, P() = 1 - P(). From the Poisson tables, I also get 1 - 0.1301 = 0.8699.
For part b) your workings are also correct: 2(0.1301 x 0.8699) = 0.2263 (4 s.f.)
I think your book meant to write P(X>7) which would make their answers right.

[Maroosh]

(Original post by magdaplaysbass)
I get the same answers as you do
It must just be an error in your book. Like you said, P() = 1 - P(). From the Poisson tables, I also get 1 - 0.1301 = 0.8699.
For part b) your workings are also correct: 2(0.1301 x 0.8699) = 0.2263 (4 s.f.)
I think your book meant to write P(X>7) which would make their answers right.

Oh thank God I found this thread or else i would have spent my evening trying to figure out what i have done wrong.