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


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.

  • $\begingroup$ Thanks!!! It a new way to look at the problem in hand :) $\endgroup$ – aimheadshots Feb 1 '16 at 5:39
  • $\begingroup$ You are welcome. Often, division into "cases" is unavoidable. But not this time. $\endgroup$ – André Nicolas Feb 1 '16 at 5:40
  • $\begingroup$ Sorry, I misinterpreted the last two lines. Very ingenious way of simplifying the problem, + lots ! $\endgroup$ – true blue anil Feb 2 '16 at 3:00
  • $\begingroup$ @trueblueanil: No problem. It would help me to know what it is about the post that was misinterpreted, so I can change anything that is unclear. $\endgroup$ – André Nicolas Feb 2 '16 at 3:11
  • $\begingroup$ I misinterpreted it as implying summation of the probabilities. Actually, your posts are the simplest and most jargon-free ones one could wish for ! $\endgroup$ – true blue anil Feb 2 '16 at 3:26

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