Example 3.29 from Statistics, 3.5 Tree and Venn Diagrams

Tree and Venn Diagrams: example 3.29

Forty percent of the students at a local college belong to a club and 50 percent work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let C = student belongs to a club and PT = student works part time.

Start by drawing a rectangle to represent the sample space. Then draw two circles or ovals inside the rectangle to represent the events of interest: belonging to a club (C) and working part time (PT). Always draw overlapping shapes to represent outcomes that are in both events.

Image shows a Venn diagram consisting of two overlapping circles inside a rectangle. The left oval is labeled C and the right is labeled PT. The rectangle is labeled S. The overlapping region is labeled C and PT and contains the probability five hundredths. The region inside C, but outside the overlapping region, contains the probability thirty-five hundredths. The region inside B, but outside the overlapping region, contains the probability forty-five hundredths. The probability fifteen hundredths is listed in the region inside the rectangle and outside the ovals.

Figure 3.17

Label each piece of the diagram clearly and note the probability or frequency of each part. Start by labeling the overlapping section first. Note that the probabilities in C total 0.40 and the sum of the probabilities in PT is 0.50. The total of all probabilities displayed must be 1, representing 100 percent of the sample space.

If a student is selected at random, find the following:

the probability that the student belongs to a club.
the probability that the student works part time.
the probability that the student belongs to a club AND works part time.
the probability that the student belongs to a club given that the student works part time.
the probability that the student belongs to a club OR works part time.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
P(C) = .40
P(PT) = .50
P(C AND PT) = .05
$P (C | P T) = \frac{P (C AND P T)}{P (P T)} = \frac{.05}{.50} = .1$
P(C OR PT) = P(C) + P(PT) − P(C AND PT) = .40 + .50 − .05 = .85

How did it go?

Solution

Solution