>>Dr. Kash Barker: The number of cracks in a section of interstate highway that are significant
enough to require repair occur with an average rate of 2.3 racks per mile. What is the probability
that there are no cracks that will require repair in the next five miles of highway? Okay, so we’ve
got a rate lambda, and that is 2.3 cracks per mile. Okay, and we want to know what's
the probability that there are no cracks being required to repair in the next five miles.
So x here is going to be a random variable that counts the number of cracks in five miles.
Okay, naturally our rate and our random variable have to agree with each other with respect to
the interval, so this lambda here is going to be five times 2.3 which is 11.5 cracks
per five miles. Okay, so our rate, 2.3 cracks per mile is equivalent to 11.5 cracks per
fives miles on average. And we want to know what’s the probability that we see no cracks in the
next five miles. Okay? So what’s the probability that our random variable takes on the realization
zero. Okay, straightforward, we’ve got a rate 11.5 to the zero, e to the minus 11.5
divided by zero factorial. And as it turns out that value is 0.1003, okay? So on average
ten percent of the five mile stretches that we might encounter will require, will have
no cracks, no cracks that need repair. Okay, we’ll move on and ask another question. What’s
the probability that at least one crack requires repair in half a mile of highway. Okay, so let’s
change up our lambda. Lambda was 2.3 cracks per mile and we need a half a mile as our
interval, so that would be half of this 1.15 cracks per half mile. Okay, and our random
variable, let’s call it y, is going to be the number of cracks requiring repair in half
a mile. Okay and so now we want to know what’s the probability that at least one crack is going
to require repair in the next half mile. That’s the probability that y is at least one. Okay,
at least one, one or more, inclusive of one, cracks in the next half mile. Ok. That is
the probability that y equals one, plus the probability that y equals two, all the way
to the probability that y equals infinity. Okay, because we are not bounded on the upper side
for this random variable. In theory we could have a million here we could have eighteen
million, whatever it might be, we don’t have an upper bound on a Poisson random variable.
You could spend a lifetime solving this problem, or we could think of it this way. What are
the possible realizations of this random variable? Zero, one, two, three, all the way up to infinity.
And we want one or more. Ok? Translating this into this kind of statement says this. So
we could solve this by taking the complement of the probability that y equals zero. So
the probability that y is greater than or equal to one is the complement of the probability
that y equals zero. That is one minus our rate which is 1.15 to the zero, e to the minus
1.15, divided by zero factorial. And if we work that out we should get a probability
of six, 0.6834. Okay? So the probability that we see one or more cracks in the next half
mile? 0.6834. On average we’ll see sixty-eight percent of one half mile stretches that have one
or more cracks.