Binomial Probably Variant help

haml

New Member
#1
Ok so what i'm thinking here is that the P(X=3) = [(2 choose 2) X (7 choose 2) ] / 10!


Am I on the right track here?
 

BGM

TS Contributor
#2
The denominator should be \( \binom {10} {5} \) instead of \( 10! \).

Note that \( 10! \) corresponding to the permutation of 10 distinct objects, while you have two groups of object in which they are indistinguishable within group.

Therefore you need to reduce the duplicated one by dividing \( (5!)^2 \) which result in the answer.

Actually the formula can be interpret as

\( \Pr\{X = x\} = \frac {\displaystyle \binom {x-1} {2} \binom {1} {1} \binom {10-x} {2}} {\displaystyle \binom {10} {5}}, ~~ x = 3, \ldots, 8 \)

and you may verify that they actually sum to 1.
 
#3
When x=3 the string has to start MMM.
When x=8 the string has to end MMM, so by symmetry P(x=3) = P(x=8)
Likewise P(x=4) = P(x=7) and P(x=5) = P(x=6)
That should save some calculations.

I'm pretty sure \(\binom{10-x}{3} \mbox{ should be} \binom{10-x}{2}\)

When x=8 the original formula gives \(\binom{2}{3}\)