The probability that out of 10 person, all born in June, at least two have the same birthday is (a) `(""^(30 C_(10)))/((30)^(10))` (b) `1 - (""^(30) C_(10))/(30!) ` (c) `((30)^(10) - ""^(30) P_(10))/((30)^(10)) ` (d) None of these

Options

A. `(""^(30 C_(10)))/((30)^(10))`
B. `1 - (""^(30) C_(10))/(30!) `
C. `((30)^(10) - ""^(30) C_(10))/((30)^(10)) `
D. None of these

Answer

To solve the problem of finding the probability that out of 10 persons all born in June, at least two have the same birthday, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the probability that at least two out of 10 people share the same birthday in June.

2. **Using the Complement Rule**:
Instead of directly calculating the probability of at least two people sharing a birthday, we can use the complement rule. The complement of "at least two people share a birthday" is "no one shares a birthday." Thus, we can express our desired probability as:
\[ P(\text{at least 2 share}) = 1 - P(\text{none share}) \]

3. **Calculating the Total Outcomes**:
In June, there are 30 days. The total number of ways to assign birthdays to 10 people (where each can have any of the 30 days) is:
\[ 30^{10} \]

4. **Calculating the Favorable Outcomes (None Share)**:
To find the number of ways in which no two people share a birthday, we can select 10 different days from the 30 available days. The number of ways to choose 10 days from 30 is given by the combination formula \( \binom{30}{10} \). After choosing the days, we can assign the 10 people to these days in \( 10! \) (10 factorial) ways. Therefore, the number of favorable outcomes where no one shares a birthday is:
\[ \binom{30}{10} \times 10! \]

5. **Putting It All Together**:
Now we can substitute the values into our probability expression:
\[ P(\text{none share}) = \frac{\binom{30}{10} \times 10!}{30^{10}} \]
Thus, the probability that at least two people share a birthday is:
\[ P(\text{at least 2 share}) = 1 - \frac{\binom{30}{10} \times 10!}{30^{10}} \]

6. **Final Expression**:
We can express \( 10! \) in terms of permutations:
\[ P(\text{at least 2 share}) = 1 - \frac{30P10}{30^{10}} \]
where \( 30P10 = \frac{30!}{(30-10)!} \).

### Conclusion:
The final probability that at least two of the 10 persons have the same birthday is given by:
\[ P(\text{at least 2 share}) = 1 - \frac{30P10}{30^{10}} \]
This matches option (c):
\[ \text{(c) } \frac{30^{10} - 30P10}{30^{10}} \]