Prove that P ( Y = n ) = p ( n ) = ( r N ) ( r − 1 N − 1 ) ( r − 2 N − 2 ) ..... ( r − n + 1 N − n + 1 ) .

Question

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Answer 1

Question

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

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Number 3b

Let X: EXP(0). suppose Y = 2x , the Y is i = 1 Select one: а. Gamma(n, ) b. Camme(n. ) C. Gamma(1,) d. Exp(ө)

1d. 1e. 1f.

Answer 2

Question

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

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Answer 3

Question

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

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Answer 4

Question

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

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Answer 5

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

Answer 6

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

Answer 7

Chapter 3, Problem 216SE

a.

To determine

Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

Answer 8

a.

Expert Solution

Explanation of Solution

Calculation:

A random variable Y is said to follow a Hypergeometric distribution, if the probability mass function of Y is,

p(y)=(ry)(N−rn−y)(Nn) , where N∈{0,1,2,...}, r∈{0,1,2,....,N} and n∈{0,1,2,...,N}.

Now, substitute y with n in the Hypergeometric formula.

That is,

P(Y=n)=p(n)=(rn)(N−nn−n)(Nn)=(rn)(Nn)=r!(n−r!)n!N!(N−n)!n!=r(r−1)(r−2)...(r−n+1)N(N−1)(N−2)....(N−n+1)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1)

Hence, it is proved that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Answer 13

c.

To determine

Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Answer 14

c.

Expert Solution

Explanation of Solution

Calculation:

The multiplication of two polynomials is defined as below:

(∑i=0maixi)(∑j=0mbjxj)=∑i=0m∑j=0naibjxi+j︸i+j=k=∑k=0m+n(∑l=0kaibk−l)xk,

where al=0 for l>m, bl>0 for all l>n.

Now, it is needed to compare the coefficient of an in (1+a)N1+N2 with that of in the expansion of (1+a)N1 and (1+a)N2.

Thus,

∑l=0n(Nl l)(N2n−l)=(N1+N2 n)(N1 0)(N2 n)+(N1 1)(N2 n−1)+.....+(N1 n)(N2 0)=(N1+N2 n)

Hence, it is prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Answer 15

c.

Expert Solution

Answer 16

c.

Expert Solution

Answer 17

Expert Solution

Answer 18

d.

To determine

Prove that ∑y=0np(y)=1.

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

Answer 19

d.

To determine

Prove that ∑y=0np(y)=1.

Answer 20

d.

Expert Solution

Explanation of Solution

Calculation:

From Part (c), it is proved that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

Thus,

∑y=0np(y)=∑y=0n(ry)(N−rn−y)(Nn)=1(Nn)∑y=0n(N1y)( N2n−y)=(N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)(Nn)=(Nn)(Nn)=1

Hence, it is proved that ∑y=0np(y)=1.

Answer 21

d.

Expert Solution

Answer 22

d.

Expert Solution

Answer 23

Expert Solution

Answer 24

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Prove that P ( Y = n ) = p ( n ) = ( r N ) ( r − 1 N − 1 ) ( r − 2 N − 2 ) ..... ( r − n + 1 N − n + 1 ) .

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Prove that P(Y=n)=p(n)=(rN)(r−1N−1)(r−2N−2).....(r−n+1N−n+1).

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Prove that (N1 0)(N2 n)+(N1 1)( N2 n−1)+....+(N1 n)(N2 0)=(N1+N2 n).

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Prove that ∑y=0np(y)=1.

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