Probability formula (not the answer)?

Binomial Distribution with n = 10, p = 0.6 and q = 0.4

P(X = 0) = 10choose0 * 0.6^0 * 0.4^10 = 0.4^10

P(X=5) = 10choose5 * 0.6^5 * 0.4^5, where 10choose5 = 10*9*8*7*6/(5*4*3*2*1)

P(X>5) = P(X=6) + P(X=7) + ... + P(X=10)

P(X = K) = 10chooseK * p^K * q^(10-K). plug & play

P(X=10) = 10choose10* 0.6^10 * 0.4^0 = 0.6^10

The binomial threat formula says P(getting ok successes in n trials) = (n opt for ok) * (p^ok) (a million-p)^(n-ok) the place p is the prospect of achievement in one trial A) here, n = 10 because of the fact he has 10 possibilities to get a ultimate question. ok = 2 because of the fact we would desire to be attentive to what occurs whilst he gets precisely 2 precise. And p = a million/5 because of the fact on any given question he has a million/5 of threat of guessing properly. So we've (10! / 2!8!) * (a million/5)^2 (4/5)^(8) B) for buying no questions, ok =0. stick to something of the formula. C) Getting a minimum of one answer ultimate ought to be a million - P(getting no solutions ultimate). So hire the respond from B. D) Getting a minimum of 9 ultimate is P(getting precisely 9 ultimate) + P(getting precisely 10 ultimate). Use the formula for each physique.