- #1
CAF123
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Homework Statement
I flip a coin 10 times and want to know the probability of getting exactly 7 heads given that the first flip is heads.
The Attempt at a Solution
The question above is part of a larger question involving the binomial RV, but I am only needing assistance here.
Using Bayes's Formula I get to the stage where I need to compute P(first one is heads| 7 heads out of 10 flips).
So what I said was there is 10 choose 7 different rearrangements of 7 heads out of 10 flips. (So this is the reduced sample space). Then I considered how many ways to get 7 heads out of 10 flips given that the first is a head. This implies there is one choice for the first flip and another 9 choose 6 ways to still attain 7 heads out of 10 flips. Therefore the probability I want is 1 x (9 choose 6)/(10 choose 7). I believe this is correct.
Someone else said this was equal to 7 x (9!)/(10!). I don't quite sure where this comes from. I think the 10! comes from the fact that you can rearrange the 10 flip outcomes 10! ways = |s|. Then if you glue the first head (flip 1) in place, then there are 9! different ways to rearrange the other flips. I don't see why this is multiplied by 7. If you consider another head out of the 7 heads, then you can't rearrange the other 9 members randomly since the first flip is fixed to be heads.
What have I missed?
Also how would I compute P(first one is heads and(intersect) we have 7 heads out of 10 flips) if I appealed to the definition of cond. prob explicitly?
thanks.