Suppose that Y is a binomial random variable with n > 2 trials and success probability p . Use the technique presented in Theorem 3.7 and the fact that E { Y ( Y − 1)( Y − 2)} = E ( Y 3 ) − 3 E ( Y 2 ) + 2 E ( Y ) to derive E ( Y 3 ).

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Answer 1

Textbook Question

Chapter 3.4, Problem 55E

Suppose that Y is a binomial random variable with n > 2 trials and success probability p. Use the technique presented in Theorem 3.7 and the fact that E{Y(Y − 1)(Y − 2)} = E(Y³) − 3E(Y²) + 2E(Y) to derive E(Y³).

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

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Answer 2

Textbook Question

Chapter 3.4, Problem 55E

Suppose that Y is a binomial random variable with n > 2 trials and success probability p. Use the technique presented in Theorem 3.7 and the fact that E{Y(Y − 1)(Y − 2)} = E(Y³) − 3E(Y²) + 2E(Y) to derive E(Y³).

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

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Answer 3

Textbook Question

Chapter 3.4, Problem 55E

Suppose that Y is a binomial random variable with n > 2 trials and success probability p. Use the technique presented in Theorem 3.7 and the fact that E{Y(Y − 1)(Y − 2)} = E(Y³) − 3E(Y²) + 2E(Y) to derive E(Y³).

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

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Answer 4

Textbook Question

Chapter 3.4, Problem 55E

Suppose that Y is a binomial random variable with n > 2 trials and success probability p. Use the technique presented in Theorem 3.7 and the fact that E{Y(Y − 1)(Y − 2)} = E(Y³) − 3E(Y²) + 2E(Y) to derive E(Y³).

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

Answer 8

Expert Solution & Answer

To determine

Derive E(Y3).

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np

Hence, the expression is E(Y3)=3n(n−1)p−n(n−1)(n−2)p3+np.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

Derive E(Y3).

Answer 12

Explanation of Solution

Calculation:

Theorem 3.7:

If Y is a binomial random variable based on n trials and success probability p, then E(Y2)=n(n−1)p+np

The random variable Y is binomial random variable having n>2 trails and probability os success p. Also, E{Y(Y−1)(Y−2)}=E(Y3)−3E(Y2)+2E(Y).

Consider the term E{Y(Y−1)(Y−2)}.

E{Y(Y−1)(Y−2)}=∑y=0n[y(y−1)(y−2)]p(y)=∑y=0n[y(y−1)(y−2)]n!y!(n−y)!py(1−p)n−y=∑y=3n[y(y−1)(y−2)]n(n−1)(n−2)(n−3)!y(y−1)(y−2)(y−3)!(n−y)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y)!py(1−p)n−y

Add and subtract 3 to the term (n−y)! in the expression.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−y−3+3)!py(1−p)n−y=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y

Suppose that z=y−3 this implies that y=z+3.

E{Y(Y−1)(Y−2)}=∑y=0nn(n−1)(n−2)(n−3)!(y−3)!(n−3−(y−3))!py(1−p)n−y=n(n−1)(n−2)∑z=0n−3(n−3)!(z)!(n−3−(z))!pz+3(1−p)n−(z+3)=n(n−1)(n−2)∑z=0n−3(n−3 z)pzp3(1−p)n−3−z=n(n−1)(n−2)p3∑z=0n−3(n−3 z)pz(1−p)n−3−z

It is obvious that, p(z)=(n−3 z)pz(1−p)n−3−z is a binomial function which is based on (n−3) number of trails. Then, ∑zp(z)=1. Thus, the value of function ∑z=0n−3(n−3 z)pz(1−p)n−3−z is 1.

That is,

E{Y(Y−1)(Y−2)}=n(n−1)(n−2)p3(1)=n(n−1)(n−2)p3

The value of E(Y3) is,

E(Y3)=3E(Y2)−E{Y(Y−1)(Y−2)}−2E(Y)=3(n(n−1)p+np)−(n(n−1)(n−2)p3)−2np [Using theorem 3.7 and E(Y)=np]=3n(n−1)p+3np−n(n−1)(n−2)p3−2np=3n(n−1)p−n(n−1)(n−2)p3+np