7 fair coins are mixed with 3 unfair coins; PMF and expected number of flips for 3 heads?

Question

7 fair coins and 3 coins with probability for heads of 0.7 are mixed. One random coin is drawn, and flipped until 3 heads are flipped. What is the probability mass function of the number of tosses needed until 3 heads are flipped, and what is the expected number of flips required for 3 heads?

I'm having difficulties writing the pmf of the number of tosses required. The probability of selecting a fair coin is 0.7, P(unfair coin) is 0.3. Furthermore, by the geometric random variable, $$ p(x)=(1-p)^{x-1} $$ so my logic is it is equal to $$ 0.7(1-0.5)^{x-1}*0.3(1-0.7)^ {x-1}$$ and the expected value is 1/p, which is $$3/(0.7*0.5)=20$$ I'm unsure whether my thought process was correct because the PMF looks different from what I have seen before.

Thanks in advance!

Answer 1

Your approach to handling the fact that there are $7$ fair coins and $3$ biased coins is partially correct, but you are using the wrong conditional distribution. They are asking about the expected number of tosses until three heads occur, not one, so the conditional distribution given which type of coin is picked is negative binomial, not geometric.

However, instead of multiplying, you need to add: $$ p(X=x) = 0.7*P(X=x|p=.5) + 0.3*P(X=x|p=.7).$$

(Also, even if it were geometric, your formula used should have been $p(x) = (1-p)^{x-1}p,$ not $p(x) = (1-p)^{x-1}$)