Print or save as PDF — or build a custom worksheet from any module's questions.
In a class of 30 students, 18 play sport and 15 play a musical instrument. Some do both. How many do neither? How would you figure this out?
A visual tool for sorting outcomes into overlapping sets — showing what is in A only, B only, both, or neither.
The most common error is filling in the total for A (say 18) straight into the left circle, rather than the "A only" value. If 18 play sport and 8 do both, then only $18 - 8 = 10$ play sport but not music. Always start with the intersection and work outward.
A Venn diagram has four regions:
All four regions must sum to $n(\xi)$, the total number of elements.
$$n(\text{A only}) + n(A \cap B) + n(\text{B only}) + n(\text{neither}) = n(\xi)$$Given: $n(\xi) = 30$, $n(A) = 18$, $n(B) = 15$, $n(A \cap B) = 8$.
Step 1: Place the intersection: put 8 in the overlap.
Step 2: A only $= n(A) - n(A \cap B) = 18 - 8 = 10$
Step 3: B only $= n(B) - n(A \cap B) = 15 - 8 = 7$
Step 4: Neither $= n(\xi) - 10 - 8 - 7 = 30 - 25 = 5$
Check: $10 + 8 + 7 + 5 = 30$ ✓
Using the example above (total = 30, regions: 10, 8, 7, 5):
When you add $P(A) + P(B)$, you count the intersection twice. So you must subtract it once:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$Verification using the example:
$$\frac{18}{30} + \frac{15}{30} - \frac{8}{30} = \frac{25}{30} = \frac{5}{6} \checkmark$$This matches the value we found by counting regions directly. The addition rule is particularly useful when you are not given all regions of the Venn diagram.
Filling a Venn diagram: start with $n(A \cap B)$, then A only $= n(A) - n(A \cap B)$, then B only $= n(B) - n(A \cap B)$, then neither $= n(\xi) - \text{all others}$.
Addition rule:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$Mutually exclusive events: $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.
In a Venn diagram, $n(\xi)=40$, $n(A)=20$, $n(B)=18$, $n(A \cap B)=6$. What is the value of "A only"?
$n(\xi)=50$, A only $= 12$, $n(A \cap B) = 5$, B only $= 18$. How many are in "neither"?
$n(\xi)=30$, $n(A \cap B)=8$. What is $P(A \cap B)$?
$P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.2$. Find $P(A \cup B)$ using the addition rule.
Two events A and B are mutually exclusive. Which statement is true?
Q6. In a group of 40 students, 22 like action movies, 17 like comedy, and 9 like both. (a) Draw and complete a Venn diagram. (b) Find $n(\text{neither})$. (c) Find $P(\text{likes action only})$.
Q7. From a completed Venn diagram where $n(\xi)=50$, A only $= 15$, $n(A \cap B) = 10$, B only $= 12$. Find: (a) $P(A \cup B)$, (b) $P(A \cap B)$, (c) $P(\text{B only})$.
Q8. Use your Venn diagram from Q6 to verify the addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Show all fractions and confirm both sides are equal.
Action only $= 22 - 9 = 13$. Comedy only $= 17 - 9 = 8$. Both $= 9$.
$n(\text{neither}) = 40 - 13 - 9 - 8 = 10$
$P(\text{action only}) = \dfrac{13}{40}$
$P(A \cup B) = \dfrac{15 + 10 + 12}{50} = \dfrac{37}{50}$
$P(A \cap B) = \dfrac{10}{50} = \dfrac{1}{5}$
$P(\text{B only}) = \dfrac{12}{50} = \dfrac{6}{25}$
$P(A) = \dfrac{22}{40}$, $P(B) = \dfrac{17}{40}$, $P(A \cap B) = \dfrac{9}{40}$
$P(A) + P(B) - P(A \cap B) = \dfrac{22+17-9}{40} = \dfrac{30}{40} = \dfrac{3}{4}$
Direct count: $\dfrac{13+9+8}{40} = \dfrac{30}{40} = \dfrac{3}{4}$ ✓
Three events A, B, C in a Venn diagram. $n(\xi)=50$, A only $= 8$, B only $= 12$, C only $= 6$, $n(A \cap B \text{ only}) = 4$, $n(B \cap C \text{ only}) = 3$, $n(A \cap C \text{ only}) = 5$, $n(A \cap B \cap C) = 2$.
(a) Find $n(\text{neither})$. (b) Find $P(\text{exactly one event occurs})$. Show all working.