Assume a day=10 hours and all 7 days of every week, you expect the same number o

Question

Assume a day=10 hours and all 7 days of every week, you expect the same number of workers to be present.

Nails fall on the ground at an average rate of 5 per day.

a) What is the probability that there are 2 days in a week where no nails fall on the ground?

b) 2 of the workers who work everyday say that they dropped a nail on the ground last week. What is the probability that they dropped them on different days?

c) What is the probability that between 2000 and 2100 (inclusive) nails fall during a 365 day year?

Explanation / Answer

Solution

Back-up Theory

Let X = Number nails that fall on the ground per day. Then, X ~ Poisson (), where = average number of nails that fall on the ground per day = 5 (given)

If a random variable X ~ Poisson(), i.e., X has Poisson Distribution with mean then

probability mass function (pmf) of X is given by P(X = x) = e – .x/(x!) …………..(1)

where x = 0, 1, 2, ……. ,

Values of p(x) for various values of and x can be obtained by using Excel Function.

If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(), then Y ~ Poisson (k) …………….. (2)

Now, to work out solution,

Part (a)

Probability that there are 2 days in a week where no nails fall on the ground.

These 2 days can be any 2 of the 7 days in a week. So, there are 7C2 combinations.

Probability that no nails fall on the ground on 2 days

= Probability that no nails fall on the ground on day1 x Probability that no nails fall on the ground on day 2 = {P(X = 0)}2

So, Probability that there are 2 days in a week where no nails fall on the ground

= (7C2) x 0.0067382 = 0.00095 ANSWER

Part (b)

Probability that the workers dropped the nails on 2 different days.

The first day can be any one of the 7 days and then the second day can only be any one of the remaining 6 days. So, there are 42 possibilities. Two workers can drop nails on the same day in 7 ways, i.e., on day1 or day2, ……., on day7.

Now, the two workers can drop on 2 different days or on the same day. Hence, there are (42 + 7) = 49 possibilities of which 42 give what we want. So,

Probability that the workers dropped the nails on 2 different days

= 42/49 = 6/7 ANSWER

Part (c)

Probability that between 2000 and 2100 (inclusive) nails fall during a 365 day year.

If Y = Number nails that fall on the ground per 365-day year, then,

[vide (2) under Back-up Theory], Y ~ Poisson (365 x 5) = Poisson (1825). So,

Probability that between 2000 and 2100 (inclusive) nails fall during a 365 day year

= P(2000 Y 2100) = P(Y 2100) - P(Y 2000) = 1 – 0.999976 = 0.000024 ANSWER

[All the above probability values have been obtained using Excel Function]

NOTE

Probability under Part (c) can also be evaluated by Normal approximation governed by:

P(2000 Y 2100) = P[{(2000 – 1825)/sqrt1825} Z {(2100 – 1825)/sqrt1825}]

= P(4.0964 Z 6.4372) = 0. 0.000022

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