Anne is attempting to take some photographs of her four children; Jeremy (aged 1

Question

Anne is attempting to take some photographs of her four children; Jeremy (aged 10), Lauren (aged 6), Nathaniel (aged 4) and Andrew (aged 2), but is having a lot of trouble trying to take photographs without them blinking. The probability that each child will blink in a photograph is 0.05, 0.07, 0.15 and 0.2 respectively, independently of each other, and independently between photos. In answering these questions, make sure to define all random variables that you use. (a) What is the probability that a group photograph (a photograph of all four children together) will have all four children not blinking? (b) If Anne attempts 20 group photographs, what is probability that at least half of them with have all four children not blinking? You may appropriately round your answer from (a) if you wish to use the Fawcett and Kent tables. (c) If Anne takes 10 individual photographs of Jeremy, what is the probability that he will have blinked in fewer than 2 of them? (d) What is the probability that Anne will need to take exactly 5 individual photographs of Lauren to have 3 photographs without her blinking? (e) What is the probability that Anne will need to take at least 3 individual photographs of Andrew until she has one where he hasn't blinked? (f) Anne wants 5 good group photos (that is, photos with no blinking) and 3 good individual photographs of each child. If each attempted photograph will cost Anne 75c to develop (Anne still uses film photography, yes everyone, you used to have to pay for photographs to be developed!), how much money will Anne expect to spend to achieve her goal?

Explanation / Answer

Part-a

As children are independent, so joint probability is product of individual probabaility

P(all not blinking)=(1-0.05)*(1-0.07)*(1-0.15)*(1-0.2)= 0.60078

Part-b

Let X denote the number of not blinking in 20 group photographs, then X follows binomial with n=20 and p=0.60078

So, P(X>=10)=1-P(X<=9)

=0.8740 using excel function =1-BINOMDIST(9,20,0.60078,TRUE)

Part-c

Let X denote the number of timed Jeremy blinks in 10 individual photographs, then X follows binomial with n=10 and p=0.05

So, P(X<2)=1-P(X<=1)

=0.0861 using excel function =1-BINOMDIST(1,10,0.05,TRUE)

Part-d

Let X denote the number of timed Lauren not blinks in 5 individual photographs, then X follows binomial with n=5 and p=1-0.07=0.93

So, P(X=3)=0.0394 using excel function =BINOMDIST(3,5,0.93,false)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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