In one town, 73% of adults have health insurance. what is the probability that 9 adults selected at random from the town all have health insurance?

You are given that in one town, 73% of adults have health insurance. You are asked to find the probability that 9 adults selected at random from the town all have health insurance.

The problem clearly made a condition that after a random sample, all 9 adults that are selected must have an insurance. Therefore, there are 9 out of 9 people in the town who has health insurance. The 73% will then be multiplied 9 times because only 73% of the adult in town has health insurance.

P (9 of 9 have health insurance) = 0.73⁹ 
P (9 of 9 have health insurance) = 0.05887
P (9 of 9 have health insurance) = 5.89%

I'll do the first one for you. it isn't very hard at all.on number one, you have (50 × 3), which equals 150, and (3 × 3), which equals 9. after you establish that, you add 150 and 9 together. so, (150 + 9) equals 159.this means that 53 × 3 = 159.