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:The probability that 4 adults selected at random from the town all have health​ insurance is 0.1897Step-by-step explanation:Consider the provided information.It is given that 66% of adults have health insurance.Let p = 66% = 0.66Therefore q=1-0.66=0.34We need to find the probability that 4 adults selected at random from the town all have health​ insurance.[tex]P(x)=^nC_r(p)^r(q)^{n-r}[/tex]Here the value of r is 4.Substitute the respective values in the above formula.[tex]P(x)=^4C_4(0.66)^4(0.34)^{4-4}[/tex][tex]P(x)=(0.66)^4(0.34)^{0}\\P(x)=0.1897[/tex]Hence, the probability that 4 adults selected at random from the town all have health​ insurance is 0.1897