a) Another binomial distribution, with Pr(claim) = 0.95 and 7 cases (n=7). So the probability of 7 women blaming their MIL out of a sample of 7 can be found using the pdf of the binomial distribution at k=7, with n=7 and p=0.95 (k=outcomes, n=trials, p=prob. each trial is the outcome).

For example using binom.dist function in Excel you get: binom.dist(7,7,0.95,0) = 0.698

b) Now k = 0, so we have binom.dist(0,7,0.95,0) = 0.000 (to three decimal places; it's a very low probability!)

c) This is equivalent to 1 minus the cumulative probability of 4 disliking their mother in law (since that cumulative probability is the sum of probabilities that 0,1,2,3 or 4 of them disk=like the MIL, so we're left with the sum of probabilities that 5,6 or 7 do). This is 1-binom.dist(4,7,0.95,1) = 0.996

d) This is the inverse of the previous question (since "more than 4" is the same as "at least 5"). So it's 0.004 to 3 decimal places (1-0.996, or binom.dist(4,7,0.95,1)).