Question 1163248
if the probability to pass the exam is .7, then the probability to fail the exam is 1 - .7 = .3
in order to not pass the exam before the fourth attempt, he would have to fail the exam the first 3 times he took it.
the probability of that happening would be .3 * .3 * .3 = .027, therefore the probability that it doesn't happen is 1 - .027 = .973.
that's the probability that he would pass the exam before the fourth attempt.
you can also look at it another way.
the probability that he would pass the exam before the fourth attempt would be equal to:
.7 if he passed it on the first attempt.
.3 * .7 if he passed it on the second attempt.
.3 * .3 * .7 if he passed it on the third attempt.
add those up and you get .7 + .21 + .063 = .973
the numbers check out.
your solution appears to be:
the probability that he would pass the exam before the fourth attempt is .973.