In one town, 67% of adults have health insurance. what is the probability the 5 adults selected at random from the town all do not have health insurance?

0.00391

Step-by-step explanation:

67% of adults have health insurance.

Adults that do not have health insurance = 100 - 67 = 33% = 0.33

Number of adults selected at random = 5

Probability that 5 adults selected at random all do not have insurance = 0.33^5

= 0.00391

P = 0.0039135

To percentage

P = 0.0039135× 100%

P = 0.39%

Therefore the probability that the 5 adults selected at random from the town all do not have health insurance is

P = 0.0039 or 0.39%

Step-by-step explanation:

Given;

Percentage of adult with health insurance = 67%

Percentage of adult without health insurance = 100% - 67% = 33%

The probability that 5 adults selected at random don't have health insurance is P.

For each selection the chances of selecting an adult without health insurance is f' = 0.33

f' = fraction of adults without health insurance.

For the five selection, the probability of selecting adults without health insurance five times is

P = (f')^5 = (0.33)^5

P = 0.0039135

To percentage

P = 0.0039135× 100%

P = 0.39%

Therefore the probability that the 5 adults selected at random from the town all do not have health insurance is

P = 0.0039 or 0.39%

26

step-by-step explanation:

c > -10.55

step-by-step explanation: