Question:
In a year of \(366\) days, what is the chance that three persons have different birthdays?
In a year of \(366\) days, what is the chance that three persons have different birthdays?
Show Hint
For birthday problems involving distinct birthdays, multiply decreasing available choices:
\[
n\times(n-1)\times(n-2)\cdots
\]
and divide by total possible outcomes.
Updated On: Jun 5, 2026
- \(\dfrac{1}{366\times366\times366}\)
- \(\dfrac{1}{366\times365\times364}\)
- \(\dfrac{365\times364}{366\times366}\)
- \(1-\dfrac{365\times364}{366\times366}\)
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The Correct Option is C
Solution and Explanation
Step 1: Find total possible birthday outcomes.
Each of the three persons can have birthday on any of the \(366\) days.
Therefore, total possible outcomes are
\[ 366\times366\times366 = 366^3 \]
Step 2: Find favorable outcomes for different birthdays.
For all three birthdays to be different:
- The first person can have birthday on any of the \(366\) days.
- The second person must have a different birthday from the first person, so there are
\[ 365 \] choices.
- The third person must differ from both first and second persons, so there are
\[ 364 \] choices.
Thus, favorable outcomes are
\[ 366\times365\times364 \]
Step 3: Compute the probability.
\[ P= \frac{366\times365\times364}{366^3} \]
Cancel one factor of \(366\):
\[ P= \frac{365\times364}{366\times366} \]
Step 4: Final conclusion.
Therefore, the probability that all three persons have different birthdays is
\[ \boxed{\frac{365\times364}{366\times366}} \]
Hence, the correct option is (C).
Each of the three persons can have birthday on any of the \(366\) days.
Therefore, total possible outcomes are
\[ 366\times366\times366 = 366^3 \]
Step 2: Find favorable outcomes for different birthdays.
For all three birthdays to be different:
- The first person can have birthday on any of the \(366\) days.
- The second person must have a different birthday from the first person, so there are
\[ 365 \] choices.
- The third person must differ from both first and second persons, so there are
\[ 364 \] choices.
Thus, favorable outcomes are
\[ 366\times365\times364 \]
Step 3: Compute the probability.
\[ P= \frac{366\times365\times364}{366^3} \]
Cancel one factor of \(366\):
\[ P= \frac{365\times364}{366\times366} \]
Step 4: Final conclusion.
Therefore, the probability that all three persons have different birthdays is
\[ \boxed{\frac{365\times364}{366\times366}} \]
Hence, the correct option is (C).
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