2) You are told that 90% of homes donate to your charitable cause. You have 46 t

Question

2) You are told that 90% of homes donate to your charitable cause. You have 46 tax receipts left and plan on canvasing a neighborhood having 50 homes (assume that each home making a donation will want a tax receipt). a) What is the expected number of homes in this neighborhood who will make a donation? b) What is the probability that you will not run out of receipts before finishing canvasing the c What is the probability that you will have receipts left over when you finish canvasing the d) A friend informs you that at least one home in the neighborhood will not make a donation. neighborhood? neighborhood? If your friend's information is correct, what now is the probability that you will have enough receipts to canvas the neighborhood?

Explanation / Answer

Solution:

2)

a) The expected number of homes in this neighbourhood who will make a donation is 45

E(x) = 0.90 × 50 = 45

b) The probability that you will not run out of receipts before finishing canvasing the neighbourhood is 0.7497.

x = 46, n = 50, p = 0.90

Binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x < 46) = 0.7497

c) The probability that you will have receipts left over when you finish canvassing the neighbourhood is 0.569.

x = 46, n = 50, p = 0.90

Binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x < 46) = 0.569

d) Now is the probability that you will have enough receipts to canvas the neighbourhood is 0.8799.

x = 46, n = 49, p = 0.90

Binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x < 46) = 0.8799

3) From the given information 12% of the wafers are defective and wafer defects occur independently

a) Let X denote number wafer inspected ubtill the first defective. Here X follows feometic distribution on with parameter p = 0.12

The p.m.f of x is given by

P(X =x) = (0.12)(0.88)x-1 , x = 1,2,3,....

The average number of wafers would be inspected before finding one which is defective is

1-p/p = 1-0.12/0.12 = 7.33

b) Probability that the 6th wafer inspected will be the first to be defective

P(X= 6) = (0.12)1 (0.88)6-1 = 0.0633

c) The probability tha the tenth wafer inspected is the second to be defective = probability that one defective wafer occur in the first 9 inspections and then 10th inspection results in the second defective.

Probability that one defective wafer occur in the first nine inspections is determined from the binomial distribution to be 9C1 (0.12)1(0.88)9-1

Because the wafer ispections are independent, the probability that 1 defective wafter occur in the first 9 inspections and the tenth inpsection on results in the second defective is the product of the probabilities of these two events,

9C1 (0.12)1(0.88)9-1 * 0.12 = 0.04

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.