Mathematics, 13.03.2020 19:53 jacquelineS4232

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?

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Answers

Answer from: KaliBratz

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

Answer from: Quest

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

Answer from: Quest

the law of definite proportions states samples of a compound will always contain the same proportion of elements by mass. in other words, a given compound will always contain the same elements in the same proportions by mass.the law of multiple proportions states when two elements combine with each other to form more than one compound, then the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.