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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.

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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.

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  • $\begingroup$ 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 !! $\endgroup$ – Math Avengers Oct 2 '18 at 21:45
  • $\begingroup$ @MathAvengers no the second term has $p^7$. The first term will have $p^8$, i.e. the value of $c=8$. $\endgroup$ – Anurag A Oct 2 '18 at 21:46
  • $\begingroup$ thank you so much ! $\endgroup$ – Math Avengers Oct 3 '18 at 3:54
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Hint: there are two possibilities for the outcome of the sixth toss. Analyse each of these cases.

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