Key takeaway: A compound event is an outcome that depends on several simple events happening, such as a die roll followed by a spin, and its chance combines the chances of the parts.
Grade 77.SP.C.8aProbability and Counting
Definition of Compound Event
A compound event asks for several things to happen together or in a row. Rolling a 6 is a simple event. Rolling a 6 and then flipping heads is a compound one, because two actions are involved.
When the actions do not affect each other, they are independent, and the chance of the combined result is the product of the separate chances. The reason is area: if one stage splits the possibilities into columns and the other into rows, the cells that satisfy both conditions form a smaller rectangle inside the whole grid. Compound events can also use “or”, which asks for either part to happen.
Formula and notation
Call the two stages \(A\) and \(B\). If neither one influences the other, the chance that both occur is a product:
\[P(A \text{ and } B) = P(A) \cdot P(B)\]
For two ways to succeed that cannot occur at once, the chances are added: \(P(A \text{ or } B) = P(A) + P(B)\).
Key facts
- Independent stages mean the first result does not change the second’s chances.
- A two-stage “and” event multiplies the chances of the stages.
- A ‘sample space’ grid or a tree lists all outcomes for checking.
- The chance of ‘and’ is never larger than the chance of either part alone.
Compound Event example
Tyrell rolls a fair die with faces 1 to 6 and then spins a wheel cut into 5 equal parts, 2 red and 3 blue. Find the chance that the die shows a 5 or a 6 and the wheel stops on red.
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Die: the faces 5 and 6 are 2 of 6 outcomes, so \(P = \frac{2}{6} = \frac{1}{3}\).
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Wheel: red covers 2 of the 5 parts, so \(P = \frac{2}{5}\).
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Neither stage affects the other, so multiply: \(\frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}\).
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Check with the grid: 4 favorable cells out of \(6 \times 5 = 30\), and \(\frac{4}{30} = \frac{2}{15}\).
The probability is \(\frac{2}{15}\), about 13.3%.
Common mistake
Watch out
Tyrell adds the chances: \(\frac{1}{3} + \frac{2}{5} = \frac{11}{15}\), and says the combined result is very likely.
Why it is wrong
“And” demands that both parts occur, which is harder than either one alone, so the chance must shrink. It should be less than \(\frac{1}{3}\), but \(\frac{11}{15}\) is bigger than both parts. Multiply instead: \(\frac{2}{15}\).
Related terms and lessons
Browse the glossary: A–Z list · Grade 7 vocabulary · Probability and Counting