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 solve the problem, we need to multiply the probability of a single adult having health insurance by itself six times since we want to find the probability that all six adults selected have health insurance.
Let the probability of an adult having health insurance be denoted as P(A). Given that 70% of adults have health insurance, P(A) = 0.70.
The probability that all six adults have health insurance, denoted as P(all six), can be calculated as follows:
P(all six) = P(A) * P(A) * P(A) * P(A) * P(A) * P(A)
P(all six) = 0.70 * 0.70 * 0.70 * 0.70 * 0.70 * 0.70
P(all six) = 0.70^6 ≈ 0.1176
Therefore, the probability that all six adults selected at random from the town have health insurance is approximately 0.1176 or 11.76%.
To find the probability that 6 adults selected at random from the town all have health insurance, we need to use the concept of independent events.
The probability that one randomly selected adult has health insurance is 70%, which can be written as a decimal as 0.70.
Since each adult's health insurance status is independent, we can multiply the probabilities together to find the probability of all 6 adults having health insurance.
P(All 6 adults have health insurance) = (0.70)^6 = 0.0279936
Therefore, the probability that 6 adults selected at random from the town all have health insurance is approximately 0.0279936 or 2.8%.