A β 1B β β[k=0..2] 15Ck(.24^k)(.76^(15-k))D β 12 and 3.1A β 0.1091B β 0.83 ...Explanation:
1. The probability is 0.52 that the officer will pull over a driver and then the expected number more. So, if x is the expected number of drivers pulled over until one is not texting, we have ...
... x = 0.52(1+x)
... 0.48x = 0.52
... x = 0.52/0.48 = 13/12 = 1 1/12 β 1 . . . . matches selection A
(Comment on this result: I find it interesting that these are the odds in favor of finding a driver who texts. That is, if the probability of texting is 0.98, the odds are 49:1 that a driver will be texting, and the expected number of pull-overs is 49.)
2. The probability of at most 2 being cured is the probability of 0, 1, or 2 being cured. You need to add up those probabilities. The sum in answer selection B does that.
3. The mean of a binomial distribution is ...
... ΞΌx = np = 60Β·0.2 = 12
... Οx = β(np(1-p)) = β(12Β·0.8) β 3.0984
These match selection D.
4. 20C14(0.8^14)(0.2^6) = 38760Β·.043980Β·0.000064 β 0.109100 . . . matches A
5. mean(x) = 0.94; mean(x^2) = 1.58, so ...
... Οx = β(1.58 -0.94Β²) β 0.83 . . . . matches selection B