Question

the probability that a randomly selected 40 - year - old male will live to be 41 years old is 0.996742, according to the national vital statistics report, vol. 71, no. 1. (a) what is the probability that two randomly selected 40 - year - old males will live to be 41 years old? (b) what is the probability that six randomly selected 40 - year - old males will live to be 41 years old? (c) what is the probability that at least one of six randomly selected 40 - year - old males will not live to be 41 years old? would it be unusual if at least one of six randomly selected 40 - year - old males did not live to be 41 years old? (a) the probability is 0.993495 (round to six decimal places as needed.) (b) the probability is (round to six decimal places as needed.)

Explanation:

Step1: <Probability of independent events>

If two events \(A\) and \(B\) are independent, the probability that both \(A\) and \(B\) occur is \(P(A\cap B)=P(A)\times P(B)\). For \(n\) independent events \(A_1,A_2,\cdots,A_n\), \(P(A_1\cap A_2\cap\cdots\cap A_n)=P(A_1)\times P(A_2)\times\cdots\times P(A_n)\). Let \(p = 0.996742\) be the probability that a 40 - year - old male will live to be 41 years old.

Step2: <Calculate for \(n = 6\) cases>

When \(n = 6\), the probability that six randomly selected 40 - year - old males will live to be 41 years old is \(P=p\times p\times p\times p\times p\times p=p^{6}\). Substitute \(p = 0.996742\) into the formula: \(P=(0.996742)^{6}\).
Using a calculator, \((0.996742)^{6}\approx0.980007\)

Answer:

\(0.980007\)