Question

In one​ town, 66% of adults have health insurance. What is the probability that 4 adults selected at random from the town all have health​ insurance? Round to the nearest thousandth if necessary.

Answer 1

Answer:

The probability that 4 adults selected at random from the town all have health​ insurance is 0.1897

Step-by-step explanation:

Consider the provided information.

It is given that 66% of adults have health insurance.

Let p = 66% = 0.66

Therefore q=1-0.66=0.34

We need to find the probability that 4 adults selected at random from the town all have health​ insurance.

$P(x)=^nC_r(p)^r(q)^{n-r}$

Here the value of r is 4.

Substitute the respective values in the above formula.

$P(x)=^4C_4(0.66)^4(0.34)^{4-4}$

$P(x)=(0.66)^4(0.34)^{0}\\P(x)=0.1897$

Hence, the probability that 4 adults selected at random from the town all have health​ insurance is 0.1897

Answer 2

Answer:

Probability that randomly selected 4  persons all have insurance is = 0.1897

Step-by-step explanation:

Given that the probability of having an insurance equals 66%=0.66

Thus

Probability that person A has insurance = 0.66

Probability that person B has insurance = 0.66

Probability that person C has insurance = 0.66

Probability that person D has insurance = 0.66

Probability that randomly selected 4  persons all have insurance is

$P(A)\times P(B)\times P(C)\times P(D)\\\\=0.66\times 0.66\times 0.66\times 0.66=0.1897$