Question

1. marital status of women according to the statistical abstract of the united states, 70.3% of females ages 20 to 24 have never been married. choose 5 young women in this age category at random. find the probability that a. none has ever been married b. at least 1 has been married source: new york times almanac.

Explanation:

Step1: Calculate the probability that a woman has never been married

The probability that a woman has never been married \(p = 0.703\), so the probability that a woman has been married \(q=1 - p=1 - 0.703 = 0.297\)

Step2: Calculate the probability for part a

We use the binomial probability formula \(P(X = k)=C(n,k)\times p^{n - k}\times q^{k}\), where \(n = 5\), \(k = 0\) (none has been married), \(C(n,k)=\frac{n!}{k!(n - k)!}\)
\(C(5,0)=\frac{5!}{0!(5 - 0)!}=1\)
\(P(X = 0)=C(5,0)\times(0.703)^{5}\times(0.297)^{0}\)
\(P(X = 0)=(0.703)^{5}\approx0.171\)

Step3: Calculate the probability for part b

The probability that at least 1 has been married is \(P(X\geq1)\)
We know that \(P(X\geq1)=1 - P(X = 0)\)
Since \(P(X = 0)\approx0.171\)
\(P(X\geq1)=1 - 0.171 = 0.829\)

Answer:

a. The probability that none has ever been married is approximately \(0.171\)
b. The probability that at least 1 has been married is approximately \(0.829\)