Answer to Question #289929 in Statistics and Probability for Daniyal

Question

In a recent year 8,073,000 male students and 10,980,000 female students were enrolled

as undergraduates. Receiving aid were 60.6% of the male students and 65.2% of the female students. Of those receiving aid, 44.8% of the males got federal aid and 50.4% of the females got federal aid. Choose 1 student at random. (Hint: Make a tree diagram.) Find the probability that the student is

A male student without aid

A male student, given that the student has aid

A female student or a student who receives federal aid

Answer 1

Total students

$$N=8073000+10980000=19053000$$

a)

$$P(M)=\dfrac{8073000}{19053000}=0.4237$$

$$P(A|M)=0.606$$

$$P(\bar{A}|M)=1-P(A|M)=1-0.606=0.394$$

$$P(M\cap \bar{A})=P(M)P(\bar{A}|M)=\dfrac{8073000}{19053000}(0.394)$$

$$\approx0.1669$$

b)

$$P(M|A)=\dfrac{P(A|M)P(M)}{P(A|M)P(M)+P(A|F)P(F)}$$

$$=\dfrac{\dfrac{8073000}{19053000}(0.606)}{\dfrac{8073000}{19053000}(0.606)+\dfrac{10980000}{19053000}(0.652)}$$

$$=\dfrac{8073(0.606)}{8073(0.606)+10980(0.652)}=0.4059$$

c)

$$P(F)=\dfrac{10980000}{19053000}=\dfrac{1220}{2117}$$

$$P(Federal|M)=0.448(0.606)(\dfrac{8073000}{19053000})$$

$$=\dfrac{243.524736}{2117}$$

$$P(F\cup Federal)=\dfrac{1220}{2117}+\dfrac{243.524736}{2117}$$

$$=\dfrac{243.524736}{2117}=0.6913$$