# Probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses.

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.

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 $$p\times \binom{5}{4}p^4(1-p) \times \binom{4}{2}p^2(1-p)^2=30p^7(1-p)^3.$$
• hi, but the format for the solution is $p^{7}$ rather than 8, is that because you have calculated the probability of 5 heads in the first 5 tosses rather than 8 tosses ? Thanks !! – Math Avengers Oct 2 '18 at 21:45
• @MathAvengers no the second term has $p^7$. The first term will have $p^8$, i.e. the value of $c=8$. – Anurag A Oct 2 '18 at 21:46