Probability from Venn Diagrams (WJEC GCSE Maths & Numeracy (Double Award): Foundation): Revision Note

Check a completed Venn diagram that the frequencies sum to the total involved or that probabilities sum to 1.

At a careers fair, 28 students were asked which types of courses they were interested in.

The results are shown in the Venn diagram below where $S$ represents science courses, $T$ represents technology courses, and $A$ represents art courses.

(a) A student is chosen at random. Find the probability that the student is interested in exactly two of the three subjects.

Answer:

The following sections of the Venn diagram represent students interested in exactly two subjects

Find the total number of students interested in exactly two subjects

12 + 4 + 2 = 18

The total number of students in the diagram is 28

Write down the probability

$\frac{18}{28}$ $\left(\mathrm{or}\frac{9}{14}\right)$

(b) Let:

• Event A be “the student is interested in Art”

• Even B be “the student is interested in Science”

Show that events A and B are not independent.

Answer:

Use the definition for independence

For independent events, $\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)$

Use the diagram to find $\mathrm{P}\left(\mathrm{A}\right)$ which is the probability of a student being interested in Art

$\mathrm{P}\left(\mathrm{A}\right)=\frac{4+6+2+1}{28}=\frac{13}{28}$

Use the diagram to find $\mathrm{P}\left(\mathrm{B}\right)$ which is the probability of a student being interested in Science

$\mathrm{P}\left(\mathrm{B}\right)=\frac{12+4+6}{28}=\frac{22}{28}$

Use the diagram to find $\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)$

This is the probability that a student is interested in both art and science

$\frac{4+6}{28}=\frac{10}{28}$

Substitute these into each side of the definition for independence, then compare

If these are independent then $\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)=\mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)$

$\mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)=\frac{13}{28}\times \frac{22}{28}=\frac{286}{784}$

$\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)=\frac{10}{28}=\frac{280}{784}$

Compare the values and make a conclusion

In this case $\mathrm{P}\left(\mathrm{A}\mathrm{and}\mathrm{B}\right)\ne \mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)$

Therefore events A and B are not independent