Question

What is the probability that at least two persons out of a group of three persons were born in the same month?
A. $\dfrac{33}{144}$
B. $\dfrac{17}{72}$
C. $\dfrac{1}{144}$
D. $\dfrac{2}{9}$

Answer 1

Hint:At first, we find the probability of any two people having the same birth month and then we find the probability of all the three people having the same birth month. We first select a month, let January. The number of ways of the first case will be,
\[\begin{align}
  & \left( \text{No of ways of Person 1 with birthday on January} \right) \\
 & \times \left( \text{No of ways of Person 2 with birthday on January} \right) \\
 & \times \left( \text{No of ways of Person 3 with birthday on January} \right) \\
 & \times \left( \text{No of months} \right)\times \left( \text{No of ways of choosing two people} \right) \\
 & =1\times 1\times 11\times 12\times {}^{3}{{C}_{2}}=396 \\
\end{align}\]
The number of ways of the second case will be,
\[\begin{align}
  & \left( \text{No of ways of Person1 with birthday on January} \right) \\
 & \times \left( \text{No of ways of Person 2 with birthday on January} \right) \\
 & \times \left( \text{No of ways of Person 3 with birthday on January} \right) \\
 & \times \left( \text{No of months} \right)\times \left( \text{No of ways of choosing three people} \right) \\
 & =1\times 1\times 1\times 12\times {}^{3}{{C}_{3}}=12 \\
\end{align}\]
The number of possible outcomes will be ${{12}^{3}}=1728$ . The probability can then be found out.

Complete step by step solution:
Let there be two events A and B. A is the event for two people having the same birth month and B is the event for three people having the same birth month.
Now, let us solve for A. We select a month for now, say it be January. The number of ways of one person having January as the birth month is $1$ and the number of ways of the second person having January as the birth month is also $1$ . For the third person to be different, his birth month should not be January but one of the other \[11\] months. Also, the number of ways of selecting two such people out of three will be ${}^{3}{{C}_{2}}=3$ . So, number of ways become,
$\Rightarrow 3\times 1\times 1\times 11=33$
This calculation holds for the other months as well. So, the total ways are,
$\Rightarrow 33\times 12=396$
Now, let us solve for B. The number of ways of one person having January as the birth month is $1$ and the number of ways of the second and the third person having January as the birth month are also $1$ each. Also, the number of ways of selecting two such people out of three will be ${}^{3}{{C}_{3}}=1$ . So, number of ways become,
$\Rightarrow 1\times 1\times 1\times 1=1$
This calculation holds for the other months as well. So, the total ways are,
$\Rightarrow 1\times 12=12$
Now, the most generalised case would be that there is no restriction on the birth months. The first person can have any one of the $12$ months and so can the other two. The number of possibilities will be,
$\Rightarrow {{12}^{3}}=1728$
The probability becomes,
$\Rightarrow P\left( A\cup B \right)=\dfrac{396+12}{1728}=\dfrac{408}{1728}=\dfrac{17}{72}$
Thus, we can conclude that the required probability is $\dfrac{17}{72}$ which is option B.

Note: We can also solve the problem in a complimentary approach. We find the number of cases where none of the people have the same birth month. This will be $12\times 11\times 10=1320$ . The probability of it will be $\dfrac{1320}{1728}=\dfrac{55}{72}$ . The required probability will be $1-\dfrac{55}{72}=\dfrac{17}{72}$ .