Question

In a class, we have 61 students of different majors. There are 23 chemistry majors (C), 12 math majors (M), 9 engineering majors (E), and 17 students who are undecided (U). Of these students, 4 of them have declared both math and engineering majors. A student will randomly be chosen to win a scholarship.
Suppose that we decide to award three scholarships. What is the probability that the first winner is a chemistry major, the second winner is a chemistry major, and the third winner is an undecided major? (A student cannot win more than one scholarship).
a. 0.3096.
b. 1.0319.
c. 0.0398.
d. 0.9604.

Answer 1

Answer:

c. 0.0398.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

To solve this question, we find each separate probability, and then multiply them.

First winner is a chemistry major

23 chemistry majors out of 61 students. So

$P(A) = \frac{23}{61}$

Second winner is a chemistry major:

Considering the first event, 22 chemistry majors out of 60 students. So

$P(B) = \frac{22}{60}$

Third winner is an undecided major;

Considering the first two events, 17 undecided out of 59 students. So

$P(C) = \frac{17}{59}$

Desired probability:

$P(A \cap B \cap B) = P(A)P(B)P(C) = \frac{23}{61} \times \frac{22}{60} \times \frac{17}{59} = \frac{23*22*17}{61*60*59} = 0.0398$

So option c.