Question

Compare P(Grade 12 | opposed with P(opposed | Grade 12). (1 point)

Answer 1

Given that the table that shows the results of the survey, you can identify that the total number of students surveyed is:

$Total=5+3+9+8+12+9+15+16+10+12+6+11=116$

Let be:

- Event A: Grade 12.

- Event B: Opposed.

You need to use the Conditional Probabilty Formula:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

• You need to find:

$P(A\cap B)$

You can identify in the table that the number of students that belong to Grade 12 and Opposed is:

$6$

Therefore:

$P(A\cap B)=\frac{6}{116}=\frac{3}{58}$

The number of students that belong to Opposed is:

$Opposed=3+12+16+6=37$

Therefore:

$P(B)=\frac{37}{116}$

Now you can determine that:

$P(A|B)=\frac{\frac{3}{58}}{\frac{37}{116}}=\frac{6}{37}$

• You need to find:

$P(B|A)$

Use:

$P(B|A)=\frac{P(A\cap B)}{P(A)}$

You can determine that:

$P(A)=\frac{12+6+11}{116}=\frac{29}{116}=\frac{1}{4}$

Finally:

$P(B|A)=\frac{\frac{3}{58}}{\frac{1}{4}}=\frac{6}{29}$

Notice that:

[tex]P(A|B)

Hence, the answer is: Third option.