Probability with Venn Diagrams

Lessons

Senior 3 · Mathematics

Probability with Venn Diagrams

3 lessons

Lesson 3 of 3

Finding probabilities of real-life events with Venn diagrams

25 min

When the regions hold probabilities

A survey of households in one town (illustrative data) found that 70% own a radio, 40% own a television and 30% own both. A household is chosen at random. What is the probability that it owns neither? Nobody has given you a count of households, but you do not need one: write probabilities in the regions instead of numbers of outcomes, and the diagram works exactly as before. The only change is that the regions now add up to 1 instead of to $n (S)$ .

Fill it from the inside out, as in unit 1: both = 0.3; radio only $= 0.7 - 0.3 = 0.4$ ; television only $= 0.4 - 0.3 = 0.1$ ; neither $= 1 - (0.4 + 0.3 + 0.1) = 0.2$ .

Definition

In a Venn diagram of probabilities, each region holds the probability of the event it shows, and the regions inside the rectangle add up to $P (S) = 1$ .

Example: in the radio and television diagram, $0.4 + 0.3 + 0.1 + 0.2 = 1$ , and the probability of "radio only" is the number in that region, 0.4.

Important

The addition rule for any two events:

P(A \cup B) = P(A) + P(B) - P(A \cap B).

For mutually exclusive events, $P(A \cap B) = 0$ , so

P(A \cup B) = P(A) + P(B).

With the complement rule, $P(\text{neither}) = 1 - P(A \cup B)$ . Any one of the four quantities $P (A)$ , $P (B)$ , $P(A \cap B)$ , $P(A \cup B)$ can be found from the other three.

Worked example 1: radio and television

For the town above, find the probability that a household chosen at random owns (a) a radio or a television, (b) neither, (c) exactly one of the two, (d) a television but no radio.

Solution

(a) $P(R \cup T) = 0.7 + 0.4 - 0.3 = 0.8$ . From the diagram: $0.4 + 0.3 + 0.1 = 0.8$ .

(b) $P\big((R \cup T)'\big) = 1 - 0.8 = 0.2$ .

(d) $P(T \cap R') = 0.1$ .

Worked example 2: a missing overlap

Of the tourists who visit Rwanda in one season (illustrative data), the probability that a tourist chosen at random visits Volcanoes National Park ( $V$ ) is 0.45, Nyungwe National Park ( $N$ ) is 0.35, and at least one of the two is 0.65. Find the probability that the tourist visits (a) both parks, (b) Volcanoes only, (c) neither park.

Solution

(a) Rearrange the addition rule: $P(V \cap N) = P(V) + P(N) - P(V \cup N) = 0.45 + 0.35 - 0.65 = 0.15$ .

(b) $P(V \cap N') = 0.45 - 0.15 = 0.30$ .

(c) $1 - 0.65 = 0.35$ . Check the four regions: $0.30 + 0.15 + 0.20 + 0.35 = 1$ ✓ (Nyungwe only $= 0.35 - 0.15 = 0.20$ ).

Worked example 3: mutually exclusive events

In a secondary school (illustrative data), the probability that a pupil chosen at random is in S1 is 0.22 and the probability that the pupil is in S2 is 0.20. Find the probability that the pupil is (a) in S1 or S2, (b) in neither.

Solution No pupil is in two classes at once, so the events are mutually exclusive and the circles do not meet.

(a) $P(\text{S1} \cup \text{S2}) = 0.22 + 0.20 = 0.42$ . (b) $1 - 0.42 = 0.58$ .

Definition

Two events are independent if the occurrence of one does not change the probability of the other. Then, and only then,

P(A \cap B) = P(A) \times P(B).

You met this rule on tree diagrams in S2. On a Venn diagram it gives a test: compute $P(A) \times P(B)$ and compare it with the probability in the overlap.

Example: for radio and television, $0.7 \times 0.4 = 0.28$ , but the overlap holds 0.3. The events are not independent: owning a radio makes owning a television slightly more likely.

Worked example 4: testing for independence

In a class of 60 pupils, 12 wear glasses ( $G$ ), 10 are left-handed ( $L$ ) and 2 do both. A pupil is chosen at random. Are $G$ and $L$ independent?

Solution $P(G) = \frac{12}{60} = \frac{1}{5}$ , $P(L) = \frac{10}{60} = \frac{1}{6}$ and $P(G \cap L) = \frac{2}{60} = \frac{1}{30}$ .

$P(G) \times P(L) = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30} = P(G \cap L)$ , so the events are independent: among the left-handed pupils the fraction with glasses, $\frac{2}{10}$ , is the same as in the whole class, $\frac{12}{60}$ .

Worked example 5: at least one, by the complement

Uwase and Mugisha try a puzzle separately, so their results are independent. The probability that Uwase solves it is 0.6 and that Mugisha solves it is 0.5. Find (a) the probability that both solve it, (b) the four regions of the Venn diagram, (c) the probability that at least one of them solves it.

Solution

(a) $P(U \cap M) = 0.6 \times 0.5 = 0.3$ .

(b) Uwase only $= 0.6 - 0.3 = 0.3$ ; Mugisha only $= 0.5 - 0.3 = 0.2$ ; neither $= 1 - (0.3 + 0.3 + 0.2) = 0.2$ .

(c) "At least one" is the complement of "neither": $1 - 0.2 = 0.8$ . Because the events are independent, "neither" can also be found directly: $P(U') \times P(M') = 0.4 \times 0.5 = 0.2$ .

Warning

Mutually exclusive is not independent. The two words describe opposite situations. If $A$ and $B$ are mutually exclusive and both have positive probability, then $P(A \cap B) = 0$ but $P(A) \times P(B) > 0$ , so they are not independent: knowing that $A$ happened tells you for certain that $B$ did not. Use the addition rule for "or"; use multiplication for "and" only after checking independence.

A three-event problem from start to finish

A survey of 80 adults in one sector (illustrative data) asked about three ways of handling money: 60 use mobile money ( $M$ ), 20 use internet banking ( $I$ ) and 29 belong to a savings group, an ikimina ( $K$ ). Also, 15 use $M$ and $I$ , 20 use $M$ and $K$ , 7 use $I$ and $K$ , and 5 use all three. Filling from the centre outwards, as in unit 1, gives the diagram.

Check: $30 + 3 + 7 + 10 + 15 + 2 + 5 = 72$ adults are inside the circles, so $80 - 72 = 8$ use none of the three.

Worked example 6: probabilities from three circles

One of the 80 adults is chosen at random. Find the probability that the adult uses (a) none of the three, (b) at least one, (c) exactly one, (d) exactly two, (e) all three, (f) mobile money and a savings group but not internet banking.

Solution $n (S) = 80$ .

(a) $\frac{8}{80} = \frac{1}{10}$ . (b) $1 - \frac{1}{10} = \frac{9}{10}$ .

(c) $\frac{30 + 3 + 7}{80} = \frac{40}{80} = \frac{1}{2}$ . (d) $\frac{10 + 15 + 2}{80} = \frac{27}{80}$ . (e) $\frac{5}{80} = \frac{1}{16}$ .

(f) $M \cap K \cap I'$ is the region holding 15: $\frac{15}{80} = \frac{3}{16}$ .

Check: $\frac{8 + 40 + 27 + 5}{80} = 1$ ✓, since every adult uses none, exactly one, exactly two or all three.

Tip

"At least one" almost always means: find "none" and subtract it from 1. "None" is one region, the outside of every circle; "at least one" is all the others, which is more work to add and easier to get wrong.

Khan Academy: Addition rule for probability. Watch the Venn diagram explain why the overlap is subtracted once, then compare it with the radio and television example above.

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Practice

$P (A) = 0.5$ , $P (B) = 0.3$ and $P(A \cap B) = 0.1$ . Draw the Venn diagram of probabilities and find $P(A \cup B)$ , $P(\text{neither})$ and $P(\text{exactly one})$ .
$A$ and $B$ are mutually exclusive, with $P (A) = 0.25$ and $P (B) = 0.4$ . Find $P(A \cup B)$ and $P(A \cap B)$ .
$P (A) = 0.4$ , $P (B) = 0.5$ and $P(A \cup B) = 0.7$ . Find $P(A \cap B)$ , and decide whether $A$ and $B$ are independent.
$A$ and $B$ are independent, with $P (A) = 0.3$ and $P (B) = 0.6$ . Find $P(A \cap B)$ and $P(A \cup B)$ .
A building has two smoke alarms that work independently. Each works with probability 0.9. Find the probability that at least one works when there is a fire.
Of 100 pupils, 45 read a newspaper, 60 listen to the radio and 25 do both. A pupil is chosen at random. Find the probability that the pupil does neither, and decide whether the two events are independent.
Use the survey of 80 adults to find the probability that an adult chosen at random (a) uses mobile money but does not belong to a savings group, (b) does not use mobile money.
Kalisa says: "If $P (A) = 0.6$ and $P (B) = 0.7$ , the events $A$ and $B$ cannot be mutually exclusive." Is he right? Find the smallest possible value of $P(A \cap B)$ .

Show answersHide answers

Regions: A only 0.4, both 0.1, B only 0.2, neither 0.3. $P(A \cup B) = 0.5 + 0.3 - 0.1 = 0.7$ ; $P(\text{neither}) = 0.3$ ; $P(\text{exactly one}) = 0.4 + 0.2 = 0.6$ .
$P(A \cup B) = 0.25 + 0.4 = 0.65$ ; $P(A \cap B) = 0$ .
$P(A \cap B) = 0.4 + 0.5 - 0.7 = 0.2$ , and $0.4 \times 0.5 = 0.2$ , so they are independent.
$P(A \cap B) = 0.3 \times 0.6 = 0.18$ ; $P(A \cup B) = 0.3 + 0.6 - 0.18 = 0.72$ .
$P(\text{neither works}) = 0.1 \times 0.1 = 0.01$ , so $P(\text{at least one}) = 1 - 0.01 = 0.99$ .
At least one: $45 + 60 - 25 = 80$ , so $P(\text{neither}) = \frac{20}{100} = 0.2$ . $0.45 \times 0.6 = 0.27 \ne 0.25$ , so the events are not independent.
(a) $M \cap K'$ : $30 + 10 = 40$ , so $\frac{40}{80} = \frac{1}{2}$ . (b) $1 - \frac{60}{80} = \frac{1}{4}$ .
Yes. Since $P(A \cup B) \le 1$ , $P(A \cap B) = P(A) + P(B) - P(A \cup B) \ge 0.6 + 0.7 - 1 = 0.3$ . The overlap is at least 0.3, so it cannot be 0.

1When the regions hold probabilities

2A three-event problem from start to finish

When the regions hold probabilities

A three-event problem from start to finish