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

Answer to part a)

Probability of all four not blinking is = (1-0.05) * (1-0.07) * (1-0.15) * (1-0.20)

Probability of all four not blinking = 0.95 * 0.93 * 0.85 *0.80 = 0.60078

.

Answer to part b)

N = 20

x = > 10

p = 0.60

.

We make use of the binomial probabiltiy in this case:

P(X>=10) = nCx * (p)x*(1-p)^(n-x)

Since this is going to be cumulative probability , we can make use of ti84 Binomcdf function

n = 20

x = 9

p = 0.60

Our aim is to find : 1 - binomcdf(9,20,0.60)

We get: 0.8725

.

Answer to part c)

Blinking chance for Jeremy is 0.05

Thus , P = 0.05

N = 10

x <2

P(x <2) = P(x=0) + P(x=1)

P(x=0) = 10C0 * (0.05)^0 * (1-0.05)^10 = 0.5987

P(x=1) = 10C1 * (0.05)^1 * (1-0.05)^9 = 0.3151

.

Thus P(x < 2) = 0.5987 + 0.3151 = 0.9139

.

Answer to part d)

Probablity of Lauren not to blink is (1-0.07) = 0.93

n = 5

x=3

P(x=3) = 5C3 * (0.93)^3 * (0.07)^2

P(x=3) = 0.0394

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