Question

The probability that out of 10 person, all born in June, at least two have the same birthday is
A.$\dfrac{{^{30}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$
B.$\dfrac{{^{30}{C_{10}}}}{{30!}}$
C.$\dfrac{{{{30}^{10}}{ - ^{30}}{P_{10}}}}{{{{\left( {30} \right)}^{10}}}}$
D.$\dfrac{{{{30}^{10}}{ - ^{30}}{C_{10}}}}{{{{\left( {30} \right)}^{10}}}}$

Answer 1

Hint: Calculate the total number of possible outcomes when all 10 persons are born in June. We will then calculate the number of possible ways if all the 10 persons have birthdays on different days. Then, we will subtract the cases when all the 10 persons have birthdays on different days from the total number of possible cases.

Complete step-by-step answer:
Each person will have a birthday on some date in June. There are 30 days in June. So, the total number of possible outcomes will be ${30^{10}}$.
We now have to find out if at least two people will have birthdays on the same date.
We first find the number of possible ways if all the 10 persons have birthdays on different days.
Suppose all the 10 persons have birthdays on different days. Then, the total number of cases will be $^{30}{C_{10}}$.
To find the probability that at least two have the same birthday, we will subtract the cases when all the 10 persons have birthdays on different days from the total number of possible cases.
Hence, total number of possibilities if at least two have the same birthday is ${30^{10}}{ - ^{30}}{C_{10}}$
Now, we will calculate the probability using the formula, $\dfrac{{{\text{number of favourable outcomes}}}}{{{\text{total number of possible outcomes}}}}$
Hence, the required probability is $\dfrac{{{{30}^{10}}{ - ^{30}}{C_{10}}}}{{{{30}^{10}}}}$
Hence, option D is correct.

Note: If we want to find that at least two have the same birthday, then we have to include all the possibilities of 2 persons, 3 persons, and so on till 10 persons have birthday on the same day. This will be a very lengthy process. Hence, we will first find when no two persons have birthdays on the same date and subtract it from all the possible outcomes.