# The probability that a randomly chosen inhabitant ( of any age ) of a particular town is in paid employment is 0.4 . A random sample of seven people is drawn from the population of the town. You...

The probability that a randomly chosen inhabitant ( of any age ) of a particular town is in paid employment is 0.4 . A random sample of seven people is drawn from the population of the town. You may assume that the number of people in paid employment in this sample has an appropriate binomial distribution.

What is the probability that exactly 3 of these people are in paid employment?

What is the probability that two or fewer of these people are in paid employment?

Give the answer to 3 decimal places please !

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### 1 Answer

40% of the population has a paid job. A random sample of 7 people is selected.

Since we are given that the probability distribution is a binomial distribution ( there are a finite number of selections, each choice has exactly 2 possibilities (in this case employed or not), the probabilities are independent and the probabilities are constant), we can use ` ``P(x=c)= ` `_nC_c(p)^c(q)^(n-c) ` where n is the sample size, p is the given probability for an event, q is the complement of the probability, and c is the number of successes in n trials.

(a) P(x=3):

`P(x=3)= ` `_7C_3(.4)^3(.6)^4=35(.4)^3(.6)^4~~.290304 `

So ` P(x=3)~~.290 `

(b) `P(x<=2) `

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

`P(x<=2)= ` ` ``_7C_0(.4)^0(.6)^7+ ` `_7C_1(.4)^1(.6)^6+ ` ` ` `_7C_2(.4)^2(.6)^5 `

`=(.6)^7+7(.4)(.6)^6+21(.4)^2(.6)^5`

`~~.419904 `

`P(x<=2)~~.420 `

Note we should not use the binomial approximation since np<5.

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