Question

What is the probability that at least two persons out of a group of three persons were born in the same month (disregard year)?

• A. $\frac{33}{144}$
• B. $\frac{17}{72}$
• C. $\frac{1}{144}$
• D. $\frac{2}{9}$

Answer 1

Answer: (B)

$\frac{17}{72}$

Let us assume that each month is equally likely as a birth month

There are ${}^{12}{P}_3$12P3 choices for the three distinct birthdays, taken in order.

And there are ${12}^3$123 choices for three birthdays altogether, taken in order.

$P\left(\text{all 3 different}\right)=\frac{{}_{12}{P}_3}{{12}^3}=\frac{1320}{1728}=\frac{55}{72}$

$P\left(\text{at least 2 match}\right)=1-\frac{55}{72}=\frac{17}{72}$

Hence, the answer is $\frac{17}{72}$.