Let A be an event, and let IA be the associated indicator random variable ( IA is 1 if A occurs, and zero if A does not occur). Similarly, let IT be the indicator of another event, B . Suppose that P(A)=p , P(B)=q , and P(A∩B)=r . Find the variance of IA−IB , in terms of p , q , r . Var(IA−IB)=

Step-by-step explanation:

Given that:

Let us consider an event is A and let the associated indicator random variable is IA

In same way, an event is B and the associated indicator random variable is IB.

Suppose that P(A) = P

P(B) = q, P(AnB) = r

Now, we find the variance of IA - IB in terms of p,q,r

Var(IA - IB) = var(IA) + var(IB) - 2cov(IA, IB)

Here, we find var(IA)

From the given data we have

similarly

Frome the given data we have

from 1,2,3 we have the variance of IA - IB is

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

answer: graph 2

step-by-step explanation: