for question number 3 ols Window Help . What is ENB, the expected number of test

Question

for question number 3

ols Window Help . What is ENB, the expected number of tests given that there have been 20 con- secutive tests without a failure? 2. The time between telephone calls at a telephone switch is an exponential random variable T with expected value 0.01. Given T0.02, . What is the conditional PDF frT 0.02(t)? . What is EITIT 0.02], the conditional expected value of T? 3. Flip an unfair coin twice. On each fip, the probability of heads equals p. Let X, equal W-X1-X2 and Y = Xi +Xy. Find the number of heads (either 0 or 1) on flip i. Let I

Explanation / Answer

Question 3:

Here first we obtain the joint probabilities distribution for X1, X2

We obtain this as:

P( X1 = 0, X2 = 0) = (1-p)2

P(X1 = 0 , X2 = 1) = (1-p) p

P(X1 = 1 , X2 = 0) = (1-p) p

P(X1 = 1 , X2 = 1) = p2

Now from the above distribution we compute the values of W and Y in each case as:

Therefore the joint PDF for W, Y here is obtained as:

P(W = Y = 0) = (1-p)2
P(W = -1, Y = 1) = p(1-p)
P(W = Y = 1) = p(1-p)
P(W = 0, Y = 2) = p2

This is the required joint PDF for W, Y that is f(w, y)

Now the conditional distribution for W given Y here is obtained as:

P(W = 0 | Y = 0) = 1
P(W = -1 | Y = 1) = P(W = 1 | Y = 1) = 0.5
P(W = 0 | Y = 2) = 1

This is the required conditional distribution of W given Y that is f(w | y)

Now the conditional distribution for Y given W here is obtained as:

P(Y = 0 | W = 0) = P(Y = 2 | W = 0) = 0.5
P(Y = 1 | W = -1 ) = 1
P(Y = 1 | W = 1 ) = 1

This is the required conditional distribution of Y given W that is f(y | w)

X1 X2 Probability W Y 0 0 (1-p)2 0 0 0 1 p(1-p) -1 1 1 0 p(1-p) 1 1 1 1 p2 0 2

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