Question

. Suppose that 10 students of a class of 50 undergraduates are chosen for a behavioral experiment. The total class consists of 15 freshmen, 18 sophomores, 10 juniors, and 7 seniors. If all students are equally likely to be picked, determine the probability of the group contains exactly seven seniors and three juniors.

Answer 1

Answer:

Approximately 0% probability of the group contains exactly seven seniors and three juniors.

Step-by-step explanation:

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

In this question, the order in which the students are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Desired outcomes:

7 seniors, from a set of 7.

3 juniors, from a set of 10. So

$D = C_{7,7} \times C_{10,3} = \frac{7!}{7!(7-7)!} \times \frac{10!}{3!(10-3)&} = 1 \times 120 = 120$

Total outcomes:

10 students froms a set of 50. so

$T = C_{50,10} = \frac{50!}{10!(50-10)!} = 10272278170$

Probability:

$p = \frac{D}{T} = \frac{120}{10272278170} \approx 0$

Approximately 0% probability of the group contains exactly seven seniors and three juniors.