Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p , where 0 < p < 1.
What is the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a,b,c,d that would match the answer $ap^{7}*(1-p)^{3}+b p^{c}(1-p)^{d}$.
I don't know how to do this question, especially the overlap part.
Best Answer
Probability of $5$ heads in the first 5 tosses and the sixth toss being a tail is $p^5(1-p)$. Now you also want to have $3$ heads in the last four tosses (because the fate of the sixth toss is already decided), the probability of that happening is $\binom{4}{3}p^3(1-p)$. So the total probability in this scenario is $$p^5(1-p) \times \binom{4}{3}p^3(1-p)=4p^8(1-p)^2.$$
Now consider the other scenario: the sixth toss is a head with probability $p$. In which case the first five tosses should have only $4$ heads. The probability of that is $\binom{5}{4}p^4(1-p)$. You also now need only $2$ heads in the last $4$ tosses. The probability of that is $\binom{4}{2}p^2(1-p)^2$. So the total probability in this scenario is $$p\times \binom{5}{4}p^4(1-p) \times \binom{4}{2}p^2(1-p)^2=30p^7(1-p)^3.$$
Hopefully you can complete it now.