Question

A certain medical test is known to detect 65% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test,

what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places?

At least 8 have the disease, rounded to four decimal places?

At most 4 have the disease, rounded to four decimal places?

b)According to government data, 6% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

b. That at most 2 of them have never been married?

c. That at least 13 of them have been married?

Answer 1

a)
`P(X=10)=(10C10)(0.65)^(10) (1-0.65)^(10-10)=0.0135`

`P(X \geq 8)=0.1757+0.0725+0.0135=0.2616`

`P(X \leq 4)=0.0000276+0.00051+0.0043+0.0212+0.0689=0.0949`

b)
`P(X=2)=(15C2)(0.06)^(2) (1-0.06)^(15-2)=0.1691`

`P(X \leq 2)=0.3953+0.3785+0.1691=0.9429`

`P(X \geq 13)=1.22x10^(-14)`

Explanation:

Part a

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".

Solution to the problem

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

`X \sim Binom(n=10, p=0.65)`

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!)`

And we want to find these probabilities:

`P(X=10)=(10C10)(0.65)^(10) (1-0.65)^(10-10)=0.0135`

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

`P(X=8)=(10C8)(0.65)^8 (1-0.65)^(10-8)=0.1757`

`P(X=9)=(10C9)(0.65)^9 (1-0.65)^(10-9)=0.0725`

`P(X=10)=(10C10)(0.65)^(10) (1-0.65)^(10-10)=0.0135`

`P(X \geq 8)=0.1757+0.0725+0.0135=0.2616`

`P(X \leq 4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)`

`P(X=0)=(10C0)(0.65)^0 (1-0.65)^(10-0)=0.0000276`

`P(X=1)=(10C1)(0.65)^1 (1-0.65)^(10-1)=0.00051`

`P(X=2)=(10C2)(0.65)^(2) (1-0.65)^(10-2)=0.0043`

`P(X=3)=(10C3)(0.65)^(3) (1-0.65)^(10-3)=0.0212`

`P(X=4)=(10C4)(0.65)^(4) (1-0.65)^(10-4)=0.0689`

`P(X \leq 4)=0.0000276+0.00051+0.0043+0.0212+0.0689=0.0949`

Part b

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

`X \sim Binom(n=15, p=0.06)`

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!)`

And we want to find these probabilities:

`P(X=2)=(15C2)(0.06)^(2) (1-0.06)^(15-2)=0.1691`

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

`P(X=0)=(15C0)(0.06)^0 (1-0.06)^(10-0)=0.3953`

`P(X=1)=(15C1)(0.06)^1 (1-0.06)^(10-1)=0.3785`

`P(X=2)=(15C2)(0.06)^(2) (1-0.06)^(10-2)=0.1691`

`P(X \leq 2)=0.3953+0.3785+0.1691=0.9429`

`P(X \geq 13)=P(X=13)+P(X=14)+P(X=15)`

`P(X=13)=(15C13)(0.06)^(13) (1-0.06)^(15-13)=1.21x10^(-14)`

`P(X=14)=(15C14)(0.06)^(14) (1-0.06)^(15-14)=1.10x10^(-16)`

`P(X=15)=(15C15)(0.06)^(15) (1-0.06)^(15-15)=4.70x10^(-19)`

`P(X \geq 13)=1.22x10^(-14)`