Question

According to government data, 20% of employed women have never been married. If 10 employed women are selected at random, what is the probability a. That exactly 2 have never been married? b. That at most 2 have never been married? c. That at least 8 have been married?

Answer 1

a)
`P(X=2) = (10C2) (0.2)^2 (1-0.2)^(10-2)= 0.302`

b)
`P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)`

`P(X=0) = (10C0) (0.2)^0 (1-0.2)^(10-0)= 0.107`

`P(X=1) = (10C1) (0.2)^1 (1-0.2)^(10-1)= 0.268`

`P(X=2) = (10C2) (0.2)^2 (1-0.2)^(10-2)= 0.302`

And replacing we got:

`P(X\leq 2) = 0.107+0.268+0.302=0.678`

c) For this case we want this probability:

`P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)`

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

`P(X=8) = (10C8) (0.8)^8 (1-0.8)^(10-8) =0.302`

`P(X=9) = (10C9) (0.8)^9 (1-0.8)^(10-9) =0.268`

`P(X=10) = (10C10) (0.8)^(10) (1-0.8)^(10-10) =0.107`

And replacing we got:

`P(X \geq 8) = 0.677`

And replacing we got:

`P(X\geq 8)=0.0000779`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

`X \sim Bin (n=10 ,p=0.2)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Solution to the problem

Let X the random variable "number of women that have never been married" , on this case we now that the distribution of the random variable is:

`X \sim Binom(n=10, p=0.2)`

Part a

We want to find this probability:

`P(X=2)`

And using the probability mass function we got:

`P(X=2) = (10C2) (0.2)^2 (1-0.2)^(10-2)= 0.302`

Part b

For this case we want this probability:

`P(X\leq 2) = P(X=0) + P(X=1) +P(X=2)`

We can find the individual probabilities and we got:

`P(X=0) = (10C0) (0.2)^0 (1-0.2)^(10-0)= 0.107`

`P(X=1) = (10C1) (0.2)^1 (1-0.2)^(10-1)= 0.268`

`P(X=2) = (10C2) (0.2)^2 (1-0.2)^(10-2)= 0.302`

And replacing we got:

`P(X\leq 2) = 0.107+0.268+0.302=0.678`

Part c

For this case we want this probability:

`P(X\geq 8) = P(X=8) + P(X=9) +P(X=10)`

But for this case the probability of success is p =1-0.2= 0.8

We can find the individual probabilities and we got:

`P(X=8) = (10C8) (0.8)^8 (1-0.8)^(10-8) =0.302`

`P(X=9) = (10C9) (0.8)^9 (1-0.8)^(10-9) =0.268`

`P(X=10) = (10C10) (0.8)^(10) (1-0.8)^(10-10) =0.107`

And replacing we got:

`P(X \geq 8) = 0.677`