On the way to work Philomena must pass through six sets of traffic lights. The lights are independent of eachother annd the probability that Philomena must stop at any particular set is 0.7. Find the probability that Philomena stops at:
c) The first three sets of lights
d) The second and sixth sets online
e) Four sets of lights including the first two
could someone show me how to work this out using calc or by hand? I know it's a binomial Q.
c) The trick here is that you're going to see it and assume you have a binomial case - but you don't. See, if
)
, then Pr(X=3) is going to give you the probability that Philomena stops at ANY 3 lights, not JUST the first three. Truth is, we don't care what she does at the last three lights, so the
=0.7^3 = 0.343)
. You can test this by checking all the different permutations of this scenario,
=Pr(YYYNNN)+Pr(YYYNNY)+Pr(YYYNYN)+Pr(YYYNYY)+Pr(YYYYNN)+Pr(YYYYNY)+Pr(YYYYYN)+Pr(YYYYYY) = 0.7^30.3^3 +0.7^40.3^2+0.7^40.3^2+0.7^50.3+0.7^40.3^2+0.7^50.3+0.7^50.3+0.7^6=0.343)
d) I'm not quite sure what they're going for by "online" - assuming they mean she's fine between 2 and 6, what you want is Pr(?YNNNNY) - the reason there's a question mark in the first one is that we don't care if she made it or not, just what she did between 2 and 6. So, once again, we can just ignore that, and we get
 = 0.7^2+0.3^3 = 0.013)
. Once again, you can double check this with all the possible scenarios.
e) Now, this is where things get interesting - we want to know the probability of her stopping at ANY FOUR. This instantly means we can use binomial stuff to find the probability. But, we also know that she's already stopped at the FIRST TWO. So, this means we get a conditional probability,
 = \frac{Pr(X=4 \cap first\: two)}{Pr(first\: two)})
. Because of this new arrangement, we have to use binomPdf very carefully. If you want to be safe, you can write out ever scenario again [Pr(YYYYNN) + Pr(YYYNYN) + Pr(YYYNNY) + Pr(YYNYYN) + Pr(YYNYNY) + Pr(YYNNYY)], and find all of them, or you can be tricky like this:
If we limit our binomial probability to just the last four events, we know that any two of those she must stop. This means we now have a binomial probability with 4 trials and probability of 0.7. Then, if we add on the

from the first two she ran, we'll get the correct probability.
So, calculating it whatever way you want, you'll get 0.1297, giving us
 = \frac{0.1297)}{Pr(first\: two)})
. Now, remember from before that the probability of stopping at the first two is just

(as we ignore the other outcomes), so our final fraction becomes
 = \frac{0.1297}{0.49}=0.2646)