The Questions
70% of all vehicles pass inspection. Assuming vehicles pass or fail independently. What is the probability:
a) exactly one of the next 3 vehicles passes
b) at most 1 of the next 3 vehicles passes
The answer to a) is .189. The way I calculated it was:
P(success) * P(fail) * P(fail) +
P(fail) * P(success) * P(fail) +
P(fail) * P(fail) * P(success) = .7*.3*.3 + .3*.7*.3 + .3*.3*.7 = .189
I summed the 3 possible permutations of 1 success and 2 failures.
For b) the answer is .216. To get that answer you take your answer to a) and add the probability of exactly 0 successes which is P(fail) * P(fail) * P(fail) => .189 + .3*.3*.3 = .216
What I don't understand is why the probability of exactly 0 successes doesn't follow the pattern of exactly 1 success. Why doesn't the "formula" work:
P(fail) * P(fail) * P(fail) +
P(fail) * P(fail) * P(fail) +
P(fail) * P(fail) * P(fail) = .3*.3*.3+.3*.3*.3+.3*.3*.3 = .081
=> .189 + .081 = .27 (not .216)
Now I'm wondering if I calculated the answer to a) the wrong way, and it was merely a coincidence that I got the right answer!