A lamp has two light bulbs
with an average lifespan of 1500 hours each.
Assuming we can model the probability of failure
of these bulbs by an exponential density function
with a mean of 1500, find the probability
that both of the lamp's bulbs fail within 600 hours.
So if we let x be equal to the number of hours
until the first light bulb fails,
and y equal the number of hours
until the second light bulb fails,
we want to find the probability
that x is less than 600 and y is less than 600.
If you take a look at our notes below,
this probability is equal to the double integral
over the region R of f of x comma y,
or for of x comma y is equal to
the product of f of x sub one,
and f of y sub one, where both of these
are the exponential density function shown here,
where in our case, mu with a mean is equal to 1500.
So again, f of x comma y is going to be equal to
f of x sub one, times f of y sub one,
where f of x sub one is going to be one divided by 1500
times e raised to the power of x divided by a mean of 1500,
times f of y sub one, which is now a function of y,
so we have 1500 times ee raised
to the power of negative y divided by 1500.
Multiplying the coefficients we have f of x comma y
equals one divided by 2,250,000,
e raised to the power of negative x divided by 1500,
e raised to the power of negative y divided by 1500.
So this is going to be f of x comma y,
and because we're looking for where x is less than 600
and y is less than 600, the region R
is going to be the square region in the xy plane.
So now we know the probability of x being less than 600
and y being less than 600, is going to be equal to,
again this double integral, where f of x comma y
is this function here, so let's write this as
one over 2,250,000 times the double integral of e
to the negative x divided by 1500,
times e to the negative y divided by 1500,
now for differential a, because our region is squared,
the order of integration doesn't really matter,
differential is equal to let's say, dx, dy,
and now for limits integration for x,
because our region is square we're going to integrate
from zero to 600, and limits of integration for y
are going to be the same from zero to 600.
Now let's go ahead and evaluate this from the next slide.
We first integrate spec to x,
notice how in order to perform mu substitution
where u is equal to negative x divided by 1500,
so differential u is equal to negative one over 1500 dx,
so if we multiply both sides by a negative 1500,
we have negative 1500, differential
u is equal to differential x or dx.
So we can think of e raised to the power of negative x
divided by 1500, differential x as
negative 1500 e to the u du,
so that would give us negative 1500 divided by 2,250,000
times the integral from zero to 600 of,
int derivative is going to be e to the u,
which is really e to the negative x divided by 1500,
and we're treating y as a constant.
And this fraction simplifies to negative one over 1500.
So these are the limits of integration for x,
so we would have e to the negative y divided by 1500 times,
when x is 600 we'll have e to the negative 600
divided by 1500, and when x is zero we have e to the zero.
This simplifies to e to the negative two-fifths,
minus one which are both constants,
so let's write this as negative one over 1500
times the quantity e to the negative two-fifths minus one,
and we still have times integral from zero to 600,
of e to the negative y divided by 1500,
and we have differential y.
So now with respect to y,
which also requires u substitution,
we already used u, let's use v
so v is equals to negative y divided by 1500,
so differential v is equal to one over 1500,
differential y, multiplying both sides again
by negative 1500, we have negative 1500 differential
v equals differential y.
So understand when we integrate we have another factor
of negative 1500, so we'd have negative one over 1500,
times the quantity e to the negative two-fifths minus one,
and we'd have an extra factor of negative 1500,
and then the derivative's going to be e to the u
which is really just e to the negative y divided by 1500,
which is really just negative y, and the derivative
is e to the v, or e to the negative y divided by 1500.
Notice how this simplifies to one,
so we have e to the negative two-fifths
minus one times again when y is equal to negative 600
we have e to the negative 600 divided by 1500,
minus when y is zero we just have e to the zero,
so notice how these two factors are going to be the same,
this comes out to the quantity e raised to the power of
negative two-fifths minus one squared,
which is the probability that we're looking for.
Again, the probability that x is less than 600
and y is less than 600 is equal to the quantity e
raised to the power of negative two-fifths
minus one squared, let's convert this
to a decimal using the calculator.
And we have open parenthesis,
second natural log brings up e,
power of negative two divided by five,
right arrow, minus one, closed parenthesis and then squared,
enter, so we get approx 0.1087.
Which as a percentage would be approx 10.87 percent.
And this is the probability
that both light bulbs will fail within 600 hours.
I hope you found this helpful.