When two things happen together, such as two dice or a coin and a spinner, a sample space diagram lists every equally likely outcome. A Venn diagram sorts a group into overlapping sets.
In a sample space diagram every cell is equally likely, so \(\text{P} = \dfrac{\text{cells in the event}}{\text{total cells}}\). In a Venn diagram, fill the overlap first, then the rest of each circle, then the outside; the probability is the region's count over the total.
Fig. 1: sample space for the total of two fair dice; 5 of the 36 equally likely cells give a total of 8
Section 01
Sample space diagrams.
In the lab, choose two dice and the rule for combining them (add, multiply, difference), then pick an event. Count the shaded cells: that number over the total number of cells is the probability.
Worked example Two dice
Two fair dice are rolled and the scores are added. Use Fig. 1 to find (a) P(total 8) (b) P(total 10 or more) (c) P(a double).
There are \(6 \times 6 = 36\) equally likely cells.
A fair coin is thrown and a fair dice is rolled. Find the probability of a head and an even number.
Outcomes: H1, H2, ..., H6, T1, ..., T6, which is \(2 \times 6 = 12\).
Head and even: H2, H4, H6, which is 3 outcomes.
\(\dfrac{3}{12} = \dfrac{1}{4}\).
Answer: \(\tfrac{1}{4}\)
Worked example Two spinners
Spinner A is numbered 1, 2, 3 and spinner B is numbered 1, 2, 3, 4; both are fair. The two numbers are multiplied. Find the probability that the product is even.
The grid has \(3 \times 4 = 12\) cells.
A product is odd only when both numbers are odd: A is 1 or 3, B is 1 or 3, giving \(2 \times 2 = 4\) odd cells.
Even cells \(= 12 - 4 = 8\), so \(\dfrac{8}{12} = \dfrac{2}{3}\).
Answer: \(\tfrac{2}{3}\)
Examiner trap
(2, 6) and (6, 2) are DIFFERENT outcomes: the first dice shows 2 in one and 6 in the other. Counting them once is why students get \(\tfrac{3}{36}\) instead of \(\tfrac{5}{36}\) for a total of 8.
Section 02
Venn diagrams with two sets.
In the lab, click a region of the Venn diagram to see its count and its probability. Always fill in the overlap first, then subtract it from each circle's total.
Worked example Filling a Venn diagram
In a class of 30, 17 play football, 12 play tennis and 5 play both. A student is chosen at random. Find the probability that the student plays (a) tennis (b) exactly one of the sports (c) neither.
Of 40 people, 22 have a cat, 18 have a dog and 7 have neither. Find the probability that a person chosen at random has both.
Let \(x\) have both. Cat only \(= 22 - x\), dog only \(= 18 - x\).
All regions add to 40: \((22 - x) + x + (18 - x) + 7 = 40\).
\(47 - x = 40\), so \(x = 7\).
\(\text{P(both)} = \dfrac{7}{40}\).
Answer: \(\tfrac{7}{40}\)
Examiner trap
Do not put 17 in the football-only region: 17 is the WHOLE football circle, which already includes the 5 who play both.
Stage 1 · Learn
Check your understanding
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Two dice are rolled. How many equally likely outcomes are in the sample space?
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A coin is thrown and a dice is rolled. How many outcomes are possible?
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In a Venn diagram, which region is \(A \cap B\)?
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Two fair dice are rolled. What is the probability that the total is 12?
Section 03
Set notation in probability.
(Extended) Venn questions may use set notation: \(A \cap B\) is "A and B" (the overlap), \(A \cup B\) is "A or B or both" (everything inside the circles), and \(A'\) is "not A".
Worked example Reading the notation (Extended)
For the football (F) and tennis (T) class above, find \(\text{P}(F \cap T)\), \(\text{P}(F \cup T)\) and \(\text{P}(F \cap T')\).
\(F \cap T\) is the overlap: \(\dfrac{5}{30} = \dfrac{1}{6}\).
\(F \cup T\) is everything in the circles: \(\dfrac{12 + 5 + 7}{30} = \dfrac{24}{30} = \dfrac{4}{5}\).
\(F \cap T'\) is in F but not in T, the football-only region: \(\dfrac{12}{30} = \dfrac{2}{5}\).