How many edges does a rectangular have

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

3 years ago

Which set of ordered pairs below is a function?

Paha777 [63]

The answer is A.

Ordered pairs that equate to a function would not have two separate points with the same x-value. In other words, every x-value must be associated with only one y-value.

5 0

3 years ago

Use the graph below to determine which year was the fourth best year for unit sales

hoa [83]

Step-by-step explanation:

i don't know that one sorry

5 0

4 years ago

What is t (cos (t)) = 0?

Tpy6a [65]

In radians:
To get 0, only one part of the multiplication needs to be zero for the whole thing to be zero. Thus we get 2 equations:
t=0, and cos(t)=0
for the first equation, it's pretty self-explanatory and sp we get 0 as one solution.
For the second equation, cos(pi/2) and cos(3pi/2) both equal 0 so both pi/2 and 3pi/2 are answers. Since there is no bound to this problem, the actual answer is:
(2Z+1)pi/2 and 0
(2Z+1) means all odd integers

6 0

3 years ago

Please can someone help me please

Tomtit [17]

You can use variables to solve this problem. Lets say that m is men, w is women, and c is children. m+w+c=266
four times as many men as children in ‘math words’ would be 4c=m
twice as many women as children would be 2c=w
what we can do now is plug those in to make everything easier with one variable
4c+2c+c=266
7c=266
c=38 now we have how many children, and we need to plug it back into what we have for women and men.
4c=m 4(38)=m m=152
2c=w 2(38)=w w=76

152 men, 76 women, and 38 children

7 0

3 years ago