The probability that a head faces up is 0.4. Peter keep flipping the coin independently until either head or tail has occurred 10 times in total.
(a) Find the probability that the coin is flipped 10 times.
(b) Find the probability that the coin is flipped 12 times.
I think I should apply binomial probability law, but I tried many times, still cannot figure out the answer, please help.
In regards to a)
Flipping 10 times means that you either get all heads in a row or all tails in a row. So the answer is $(0.4)^{10} + (0.6)^{10}$
In regards to b)
$$\binom{11}{9}(0.6)^{10}(0.4)^{2} + \binom{11}{9}(0.4)^{10}(0.6)^{2}$$
The reasoning is you know the outcome of the last flip. In the case you get $10$ heads the last flip must be heads. So that means you have $11$ spots to arrange $9$ heads. Then you just add the case where you win with tails.
Word of caution
I am a beginner so I could be wrong. See if this makes sense to you.