3. When selecting a random usb flash drive from a factory, the probability of it

Question

3. When selecting a random usb flash drive from a factory, the probability of it failing is .024. Answer the following. a) What is the probability that the third failing drive selected is chosen on the 15th choice? b) What is the mean and variance of the number of selections before the third failing drive is selected? c) What is the probability that the first drive that fails is picked on the 9th selection? d) What is the mean and variance of the number of selections before the first failing drive is selected?

Explanation / Answer

p = 0.024

a) P(3rd failing drive selected is chosen on 15th choice) = P(2 failing drive out of 14) * P(failing drive)

                                                    = 14C2 * 0.242 * (1 - 0.024)12 * 0.024

                                                    = 0.094

b) thisi is negative binomial distribution

p = 0.024

r = 3

mean = r/p = 3/0.0.24 = 125

variance = pr/(1-p)2 = 0.024 * 3 / (1 - 0.024)2 = 0.076

c) P(first drive that fails is on 9th pick) = (1 - p)8 * p = (1 - 0.024)8 * 0.024 = 0.0198

d) this is geometric distribution

p = 0.024

mean = 1/p = 1/0.024 = 25 = 41.67

variance = p / (1 - p)2 = 0.024 / (1 - 0.024)2 = 0.0252

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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