In one town, 70% of adults have health insurance. What is the probability that 6 adults selected at random from the town all have health
insurance?
To find the probability that 6 adults selected at random from the town all have health insurance, we need to multiply the probability of each individual having health insurance.
The probability of an individual having health insurance is 70% or 0.70 in decimal form.
Since we are selecting 6 adults at random, we need to multiply the probability of each event happening together. Therefore, the probability that 6 adults selected at random from the town all have health insurance is:
0.70 * 0.70 * 0.70 * 0.70 * 0.70 * 0.70 = 0.027993 (rounded to 5 decimal places)
So, the probability that 6 adults selected at random from the town all have health insurance is approximately 0.028 or 2.8%.
To find the probability that 6 adults selected at random from the town all have health insurance, we need to use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successes.
nCx is the combination of n items taken x at a time.
p is the probability of success on a single trial.
n is the total number of trials.
In this case:
x = 6 (we want all 6 adults to have health insurance)
nCx = 6C6 = 1 (the combination of 6 items taken 6 at a time is 1)
p = 0.7 (the probability of having health insurance)
n = 6 (the total number of trials, in this case, the number of adults)
Let's plug in the values and calculate the probability:
P(6) = (6C6) * 0.7^6 * (1-0.7)^(6-6)
P(6) = (1) * 0.7^6 * (1-0.7)^(0)
P(6) = 1 * 0.7^6 * 1
P(6) = 0.7^6
P(6) ≈ 0.117649
So, the probability that 6 adults selected at random from the town all have health insurance is approximately 0.117649, or 11.76%.