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# Probability watch

1. AA
2. They're not equivalent.

3. (Original post by ttoby)
They're not equivalent.

Thanks for your reply! But according to the answer I have here they are equivalent, I'm not sure how they do it though. I don't think there is a mistake in the answer since its a mark scheme for an exam.
4. I agree with ttoby. I'd suggest posting a photo of the actual question.
5. (Original post by ghostwalker)
I agree with ttoby. I'd suggest posting a photo of the actual question.
Here's the question:

I got up to the summation from k=1 to infinity of (1/2)^k (11/12)^(k-1)
EDIT: Got rid of the question, mind if I pm you the question instead? Just found out I can't make the question publicly available...
EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
6. (Original post by JBKProductions)
EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
Setting aside the factor of 1/2 you have a geometric progression. First term is? Common ratio is? Apply forumla a/(1-r) = result. Then divide by 2.

Edit: I see the 1/2 has gone so ignore the bit about 1/2 and divide by 2
7. (Original post by JBKProductions)
Here's the question:

I got up to the summation from k=1 to infinity of (1/2)^k (11/12)^(k-1)
EDIT: Got rid of the question, mind if I pm you the question instead? Just found out I can't make the question publicly available...
EDIT2: Nevermind got it sorted now, but have another question about this! How do I simplify the summation from k=1 to infinity of (11/14)^k-1?
It's the sum of an infinite geometric series.
8. (Original post by ghostwalker)
Setting aside the factor of 1/2 you have a geometric progression. First term is? Common ratio is? Apply forumla a/(1-r) = result. Then divide by 2.
Thank you!
9. (Original post by ttoby)
It's the sum of an infinite geometric series.
Thank you too!

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