00:01
In this problem we are provided with the joint density function f of x comma y equals to k times x squared when 0 is less than or equal to x is less than 2 and 0 is less than or equal to y is less than or equal to 1 and it equals to 0 otherwise in subpart a we are asked to find out the value of k we know that double integral from 0 to 1 0 0 to 2, k x squared d x, d x, dy must be equal to 1.
00:39
So here since both the integrals are independent of each other, we can write them as a product of 2 integrals.
00:47
And evaluating each integral, we have y with lower limit 0 and upper limit 1, k times x cubed over 3 with lower limit 0 and upper limit 2 and this equals to 1.
01:05
So now we need to further solve this.
01:08
Substituting the upper limit and lower limit, we have 1 minus 0 times k times 2 cute which is 8 over 3 minus 0.
01:18
This equals to 1.
01:20
So further simplifying this, we have k times 8 over 3 equals to 1 which implies that the value of k is 3 over 8.
01:30
Therefore this is the required answer for subpart a.
01:35
Next, in subpart b, we are asked to find the probability, the point x comma y satisfies x plus y is less than or equal to 2.
01:53
So the probability of x plus y less than or equal to 2 equals to integral 0 to 2, integral 0 to 2 minus x, the value of k which is 3 over 8 times x.
02:09
Squared, d ,y, dx.
02:12
So first evaluating the inner integral, we have 3 over 8 times integral 0 to 2, x squared times the integral of dy which is y with lower limit 0 and upper limit 2 minus x d x.
02:27
Now evaluating the next integral but before that we need to substitute the upper and lower limit.
02:33
So we get x squared times 2 minus x.
02:36
Simplifying the integrand, we have integral 0 to 2, 2x squared minus x cubed dx.
02:45
Now evaluating this, we get 3 over 8 times 2 times x cubed over 3 minus x raised to the power 4 over 4 with lower limit 0 and upper limit 2.
02:57
So substituting the upper and the lower limits we have 3 over 8 times 2 times 2 cubed.
03:07
That is 2 times 8 over 3 minus 2 raised to the power 4 which is 16 over 4 and substituting the low limit both terms become 0...