Question

1. student financial aid 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. a male student without aid
b. a male student, given that the student has aid
c. a female student or a student who receives federal aid

Explanation:

Step1: Calculate total number of students

Total number of students \(N=8073000 + 10980000=19053000\)

Step2: Calculate part (a)

The proportion of male students is \(P(\text{Male})=\frac{8073000}{19053000}\)
The proportion of male students without aid: \(P(\text{Male without aid})=P(\text{Male})\times(1 - 0.606)\)

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Step3: Calculate part (b)

Number of male students with aid \(n_{M - aid}=8073000\times0.606 = 4892238\)
Number of female students with aid \(n_{F - aid}=10980000\times0.652=7168960\)
Total number of students with aid \(n_{aid}=4892238+7168960 = 12061198\)
By Bayes' theorem \(P(\text{Male}|\text{aid})=\frac{n_{M - aid}}{n_{aid}}\)

$$ P(\text{Male}|\text{aid})=\frac{4892238}{12061198}\approx0.405 $$

Step4: Calculate part (c)

Number of female students \(n_F = 10980000\)
Number of students with federal aid:
Number of male students with federal aid \(n_{M - federal}=4892238\times0.448 = 2201722.624\)
Number of female students with federal aid \(n_{F - federal}=7168960\times0.504=3613155.84\)
Number of students with federal aid \(n_{federal}=2201722.624 + 3613155.84=5814878.464\)
Number of female students with federal aid \(n_{F\cap federal}=3613155.84\)
By the addition rule \(P(F\cup\text{federal})=P(F)+P(\text{federal})-P(F\cap\text{federal})\)

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$$ P(F\cup\text{federal})=0.576 + 0.305- 0.190=0.691 $$

Answer:

a. Approximately \(0.167\)
b. Approximately \(0.405\)
c. Approximately \(0.691\)