Question

In our class, there are 14 female and 11 male students. a. Pick one student randomly, what is the probability you get a female student? b. Pick two students randomly, what is the probability that both are female? c. Pick three students randomly, what is the probability that you get 2 female and one male? d. Pick four students randomly, what is the probability that you get 1 female and 3 male?

Answer 1

Answer:

a) 56% probability you get a female student.

b) 30.33% probability that both are female

c) 43.52% probability that you get 2 female and one male

d) 18.26% probability that you get 1 female and 3 male

Step-by-step explanation:

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

The order in whih the students are chosen is not important. So 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.

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

a. Pick one student randomly, what is the probability you get a female student?

Desired outcomes:

1 female student from a set of 14. So

$D = C_{14,1} = \frac{14!}{1!13!} = 14$

Total outcomes:

1 student from a set of 25. So

$T = C_{25,1} = \frac{25!}{1!24!} = 25$

Probability:

$P = \frac{D}{T} = \frac{14}{25} = 0.56$

56% probability you get a female student.

b. Pick two students randomly, what is the probability that both are female?

Desired outcomes:

2 female students, from a set of 14. So

$D = C_{14,2} = \frac{14!}{2!12!} = 91$

Total outcomes:

2 students from a set of 25. So

$T = C_{25,2} = \frac{25!}{2!23!} = 300$

Probability:

$P = \frac{D}{T} = \frac{91}{300} = 0.3033$

30.33% probability that both are female

c. Pick three students randomly, what is the probability that you get 2 female and one male?

Desired outcomes:

2 female students, from a set of 14, and one male student, from a set of 11. So

$D = C_{14,2}*C_{11,1} = \frac{14!}{2!12!}*\frac{11!}{1!10!} = 1001$

Total outcomes:

3 students from a set of 25. So

$T = C_{25,3} = \frac{25!}{3!22!} = 2300$

Probability:

$P = \frac{D}{T} = \frac{1001}{2300} = 0.4352$

43.52% probability that you get 2 female and one male

d. Pick four students randomly, what is the probability that you get 1 female and 3 male?

Desired outcomes:

1 female student, from a set of 14, and 3 male students, from a set of 11.

$D = C_{14,1}*C_{11,3} = \frac{14!}{1!13!}*\frac{11!}{8!3!} = 2310$

Total outcomes:

4 students from a set of 25. So

$T = C_{25,4} = \frac{25!}{4!21!} = 12650$

Probability:

$P = \frac{D}{T} = \frac{2310}{12650} = 0.1826$

18.26% probability that you get 1 female and 3 male