Question

In a year of \(366\) days, what is the chance that three persons have different birthdays?

• A. \(\dfrac{1}{366\times366\times366}\)
• B. \(\dfrac{1}{366\times365\times364}\)
• C. \(\dfrac{365\times364}{366\times366}\)
• D. \(1-\dfrac{365\times364}{366\times366}\)

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.

Answer 1

The Correct Option is C

Step 1: Count all the ways.
Each of the three people can be born on any of the $366$ days, so the total number of birthday combinations is
\[ 366\times366\times366 \]

Step 2: Count the good cases.
For all three to differ, the first person is free with $366$ choices. The second must avoid one day, leaving $365$. The third must avoid two days, leaving $364$. So the good cases are
\[ 366\times365\times364 \]

Step 3: Form the probability.
\[ P=\frac{366\times365\times364}{366\times366\times366} \]

Step 4: Cancel one factor.
One $366$ cancels from top and bottom, giving
\[ P=\frac{365\times364}{366\times366} \]
\[ \boxed{\dfrac{365\times364}{366\times366}} \]