what is the probability that you attempt to play it at least 8 times in order to play it correctly 4 times before quitting?

Question

Question: You are working on a difficult passage from a new piece you are learning on the piano.You wish to play it correctly 4 times before calling it a day. If you have a probability of 2 of playing it correctly on every attempt, and the attempts are independent (unfortunately!), what is the probability that you attempt to play it at least 8 times?

I know my solution is incorrect because the probabilities sum over 1 but can someone fill me in on where my approach went wrong and/or how I can adjust my solution?

Thank you]1

Answer 1

For an example of where your approach goes wrong, consider your second-to-last case, where you calculate the probability that you attempt to play exactly $5$ times. If you succeed in the first $4$ times, you won't play the $5$th time because you are already finished -- you need to calculate the probability that the $5$th time is needed (and you end exactly with the $5$th time). So the $\dbinom{5}{4}$ term you have there is not correct. Does this make sense?

Answer 2

Your approach went wrong in that you count cases that can't happen. For instance, your first line $\binom 74(2/3)^4(1/3)^3$ will, among others, count the possibility of first playing correctly 4 times, then incorrectly 3 times. But that will never happen.

Instead of calculating the probability of "four correct out of seven", that line should calculate the probability of "three correct out of the first six, and then correctly on the seventh time". And similarly for the other lines of your calculation.