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
Best Answer
You set up your calculation well, but it is a little lengthy. There is a quicker way.
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.