a) Another binomial distribution, with Pr(pass) = 0.85 and 11 students (n=11). So the probability of 11 students passing out of a sample of 11 can be found using the pdf of the binomial distribution at k=11, with n=11 and p=0.85 (k=outcomes, n=trials, p=prob. each trial is the outcome).

For example using binom.dist function in Excel you get: binom.dist(11,11,0.85,0) = 0.167

b) This is equivalent to 1 minus the cumulative probability of 5 passing (since more than half = 6 or more passing, and cumulative probabilities tell us the total probability of up to k outcomes). So we can get this using

1-binom.dist(5,11,0.85,1) = 0.997

c) This is the inverse of the previous question! So it's simple 1-0.997 = 0.003, or directly, binom.dist(5,11,0.85,1).

d) The probability of all students failing is given by setting k=0: So we have binom.dist(0,11,0.85,0) = 0.000 (to 3 decimal places - it's a very tiny chance, more than 1 billion-to-one against).