3.5 Venn diagrams

This is a simple task of finding the value where the two characteristics intersect on the table, and then applying the postulate of probability, which states that the probability of an event is the proportion of outcomes that match the event in which we are interested as a proportion of all total possible outcomes.

P ( College Grad $\cap$ Female ) = $\frac{30}{200}=0.15$

This task involves the use of the addition rule to solve for this probability.

P ( College Grad $\cup$ Female ) = P ( F ) + P ( CG )− P ( P $\cap$ CG )

P ( College Grad $\cup$ Female ) = $\frac{80}{200}+\frac{90}{200}-\frac{30}{200}=\frac{140}{200}=0.70$

Here we must use the conditional probability rule (the modified multiplication rule) to solve for this probability.

P ( HS Grad $|$ Male = $\frac{P(\text{HS Grad}\cap \text{Male})}{\text{P}\left(\text{Male}\right)}=\frac{\left(\frac{15}{200}\right)}{\left(\frac{120}{200}\right)}=\frac{15}{120}=0.125$

There are two ways to approach this test. The first method seeks to test if the intersection of two events equals the product of the events separately remembering that if two events are independent than P ( A )* P ( B ) = P ( A $\cap$ B ). For simplicity's sake, we can use calculated values from above.

Does P(College Grad $\cap$ Female) = P(CG) ⋅ P(F)?

$\frac{30}{200}\ne \frac{90}{200}\cdot \frac{80}{200}$ because 0.15 ≠ 0.18.

Therefore, gender and education here are not independent.

The second method is to test if the conditional probability of A given B is equal to the probability of A . Again for simplicity, we can use an already calculated value from above.

Does P(HS Grad $|$ Male) = P(HS Grad)?

$\frac{15}{120}\ne \frac{27}{200}$ because 0.125 ≠ 0.135.

Therefore, again gender and education here are not independent.