We have a biased coin such that $P(head) = 0.25$ and $P(Tails) = 0.75$, suppose we flip coin until we observe third heads. What is the probability that we stop flipping after exactly ten flips?.
I thought that let us fix of getting a third head at last that is at 10th flip, so that we would stop there, and the remaining - getting two heads can be accommodated in the 9 trials. so there are $9$ choose 2 ways of getting two heads so the probability that we stop flipping after exactly ten flips is $^9C_{2}$ . $\frac{1}{4}^3$.$\frac{3}{4}^7$. Is this correct?
EDIT - Now the probability of getting exactly 3 heads? I got it to be $^{10} C_{3} \frac{1}{4}^3 \frac{3}{4}^7$. Should we get the same as the previous one? any reason why they should/should not be same?
$$\sum\limits_{k=3}^{10} {10 \choose k} \left( {1 \over 4} \right)^k \left( {3 \over 4} \right)^{10-k} = 0.474$$