# Revised GRE Math Probability section

The table shows the distribution of a group of $40$ college students by gender and class $$\begin{array}{c|lcr} & \text{Sophomores} & \text{Juniors} & \text{Seniors} \\ \hline \text{Males} & 6 & 10 & 2 \\ \text{Females} & 10 & 9 & 3 \\ \end{array}$$

If one student is randomly selected from this group, find the probability that the student chosen is (i) a female or a sophomore?

• How did you approach? – user67773 May 8 '13 at 7:49

How many students are either females or sophomores (or both)? First count the number of females, then count the number of male sophomores (female sophomores need not be counted since you already counted them when you counted the number of females). What percent of the total number of students is this?

There are $10 + 9 + 3 = 22$ females and $6$ male sophomores, so there are $22 + 6 = 28$ students out of $40$ who are either female or sophomore.

$28/40 = 0.7$

This question uses the inclusion exclusion principle:

$\Pr(F \cup S) = \Pr(F) + \Pr(S) - \Pr(F \cap S)\tag{1}$

$\cup$ is the math notation for "union" which is equivalent to "OR"

$\cap$ is the math notation for "intersection" which is equivalent to "AND"

Pr(F) = Probability of Females

Pr(S) = Probability of Sophomores

$\Pr(F \cap S)$ = Probability of Females and Sophomores

There are 22 females, 16 sophomores, 10 females who are sophomores, and 40 students total. So plug and chug baby!

$\Pr(F \cup S) = \cfrac{22+16-10}{40}=0.7$

You can think of the equation (1) as a venn diagram. How many elements do you have in all the circles? Its the one you have in A + B minus the elements in A + B!.

• Surely you don't mean $10+16-10$. There are $22$ total females, not $10$. – Potato May 21 '13 at 5:36
• @Potato: Thanks. Fixed it. But you catch my drift ;) – user1527227 May 21 '13 at 6:01
The number of sophomores= 16


Number of females=10+9+3=22 Number of female sophomores=6

We get a venn diagram between sophomores and females, with common area as 6 Total people who are sophomore or female= 10+6+16=32

Probability=32/40=0.8

• Even though you pointed out that the two groups have an intersection, I think you are overcounting. – luke May 8 '13 at 16:20
• And there are 10 people, not 6, in the intersection. – manthanomen May 8 '13 at 16:29
• You're solution is wrong. – user1527227 May 21 '13 at 5:22