# 15 coin toss ,coins which shows up heads is then tossed again, What is the probability of observing 5 heads in the second round of tosses?

In an experiment, n coins are tossed, with each one showing up heads with probability p independently of the others. Each of the coins which shows up heads is then tossed again. What is the probability of observing 5 heads in the second round of tosses, if we toss 15 coins in the first round and p = 0.4? (Hint: First find the mass function of the number of heads observed in the second round.)

While solving this question i calculated the all the scenario where i get heads > 5 in first toss and then I solved for 5 heads in first round of toss and then i multiplied it with the probability of getting 5 heads in the second round in each of the scenario eg :

Case 1 : 1st toss (5H,10T) 2nd toss (5H,0T) = 15C5*(0.4^5)*(0.6^10) * 5C0*(0.4^5)(0.6^0)

Case 2 : 1st toss (6H,9T) 2nd toss (5H,1T) = 15C6*(0.4^6)*(0.6^9) * 6C5*(0.4^5)(0.6^1)

Case 3 :1st toss (7H,8T) 2nd toss (5H,2T) = 15C7*(0.4^7)*(.6^8) * 7C5*(0.4^5)(0.6^2)

Similarly i exhausted all the cases till

Case 11 : 1st toss (15H,0T) 2nd toss (5H,10T) = 15C15*(0.4^15)*(.6^0) * 15C5*(.4^5)(.6^10)

then i calculated the mass function as $\sum_{n=5}^{15} (15Cn*nC5*(0.4^{n+5} * 0.6^{10}))$

Am i doing it the right way or is there any simpler way to do so?

Doing Probability after 4 years. So i m a bit rusty Thanks in advance

The coins that did not land heads in the first round will feel bad about being left out of the second round. So let's change the game a little, and toss the first coin twice, and the second, and the third, and so on. The probability a coin lands Head then Head is $(0.4)^2$, so the probability this happens $5$ times is $\binom{15}{5}((0.4)^2)^5(1-(0.4)^2)^{10}$.
Let random variable $X$ be the number of heads in the second round. You can find the distribution of $X$, that is, $\Pr(X=k)$, in the same way.