Answer to Question #205142 in Statistics and Probability for Shema Derrick

Question

.In 1970, 11% of Americans completed four years of college, 43% of them were women. In 1990, 22% of Americans completed four years of college; 53% of them were women. (Time, Jan.19, 1996). a) Given that a person completed four years of college in 1970, what is the probability that the person was a woman? b) What is the probability that a woman would finish four years of college in 1990? c) What is the probability that in 1990 a man would not finish college?

Answer 1

S= randomly chosen American completed four years of college in 1970s

N=randomly chosen American completed four years of college in 1990s

W= randomly chosen American who completed four years of college is a woman

M=randomly chosen American who completed four years of college is a man

P(S)=0.11

P(W|S)=0.43

P(N)=0.22

P(W|N)=0.53

a) P(W|S)=0.43

b)

$P(W|N) = \frac{P(W \cap N)}{P(N)} \\ 0.53 = \frac{P(W \cap N)}{0.22} \\ P(W \cap N) = 0.53 \times 0.22 = 0.1166$

c)

$(M \cap N)$ = man finished college in 1990s

$(M \cap N)’$ = man had not finished college in 1990s

The events $(M \cap N)$ and $(M \cap N)’$ are complementary:

$P(M \cap N)’ = 1 -(M \cap N) \\ P(M|N)= 0.47 \\ P(M|N)= \frac{P(M \cap N)}{P(N)} \\ 0.47 = \frac{P(M \cap N)}{0.22} \\ P(M \cap N) = 0.47 \times 0.22= 0.1034 \\ P(M \cap N)’ = 1 -0.1034 \\ = 0.8966$