a) OK, this is a binomial distribution, with Pr(extrovert) = 0.65 and 15 cases (n=15). So the probability of 10 or more extroverts in a sample of 15 can be found using 1 minus the cdf of the binomial distribution at k=9 - since that is 1 minus the total probability of 9 or fewer extroverts in this group. So n=15, k=9 and p=0.65 (k=outcomes, n=trials, p=prob. each trial is the outcome).
For example using binom.dist function in Excel you get: 1-binom.dist(9,15,0.65,1) = 1-0.436 = 0.564.
The second part is identical, but now we set k=4 ("below 5") instead of k=9: 1-binom.dist(4,15,0.65,1) = 1-0.003 = 0.997.
Finally, the probability that all are extroverts is given by the pdf at k=15: binom.dist(15,15,0.65,0) = 0.002 (to 3 decimal places).
b) Now we have Pr(introvert) = 70%, so p=0.7. And we just have 5 programmers, so n=5. The probability of none being introverts is the pdf at k=0: binom.dist(0,5,0.7,0) = 0.002.
The probability of 3 or more is 1 minus the probability of 2 or fewer (i.e. 1 minus the cdf of the binomial distribution at 2): 1-binom.dist(2,5,0.7,1) = 1-0.163 = 0.837.
The probability of all being introverted is simply the pdf at k=5: binom.dist(5,5,0.7,0) = 0.168.
Hope this helped! If you can't use the Excel functions I've given, try googling an online calculator for the binomial distribution - if you plug in the correct values for k, n and p each time you should get identical answers. Be careful and remember sometimes you are using the cdf (total sum of probabilities for a range) and other times you might use the pdf (probability for one exact outcome).
Content will be erased after question is completed.