For n independent events `A_i's, p(A_i)=1/(1+i),i=1,2,...n`. The probability that at least one of the events occurs, is

Options

A. `1//n`
B. `1// (n + 1)`
C. `n// (n + 1)`
D. None of these

Answer

To solve the problem, we need to find the probability that at least one of the independent events \( A_i \) occurs, given that \( p(A_i) = \frac{1}{1+i} \) for \( i = 1, 2, \ldots, n \).

### Step-by-Step Solution:

1. **Identify the probabilities of each event:**
- For \( i = 1 \):
\[ p(A_1) = \frac{1}{1+1} = \frac{1}{2} \]
- For \( i = 2 \):
\[ p(A_2) = \frac{1}{1+2} = \frac{1}{3} \]
- For \( i = 3 \):
\[ p(A_3) = \frac{1}{1+3} = \frac{1}{4} \]
- Continuing this pattern, we find:
\[ p(A_n) = \frac{1}{1+n} \]

2. **Calculate the probabilities of the complements of each event:**
- The complement of \( A_i \) is denoted as \( A_i' \) (not \( A_i \)):
\[ p(A_1') = 1 - p(A_1) = 1 - \frac{1}{2} = \frac{1}{2} \]
\[ p(A_2') = 1 - p(A_2) = 1 - \frac{1}{3} = \frac{2}{3} \]
\[ p(A_3') = 1 - p(A_3) = 1 - \frac{1}{4} = \frac{3}{4} \]
- In general:
\[ p(A_i') = 1 - \frac{1}{1+i} = \frac{i}{1+i} \]

3. **Find the probability that none of the events occur:**
- Since the events are independent, the probability that none of the events occur is the product of their complements:
\[ p(A_1' \cap A_2' \cap A_3' \cap \ldots \cap A_n') = p(A_1') \cdot p(A_2') \cdot p(A_3') \cdots p(A_n') \]
- Substituting the values:
\[ p(A_1' \cap A_2' \cap A_3' \cap \ldots \cap A_n') = \left(\frac{1}{2}\right) \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{3}{4}\right) \cdots \left(\frac{n}{n+1}\right) \]

4. **Simplify the product:**
- The product can be expressed as:
\[ = \frac{1 \cdot 2 \cdot 3 \cdots n}{2 \cdot 3 \cdot 4 \cdots (n+1)} \]
- Notice that the numerator is \( n! \) and the denominator can be rewritten as:
\[ = 2 \cdot 3 \cdots (n+1) = \frac{(n+1)!}{1} \]
- Therefore:
\[ = \frac{n!}{(n+1)!} = \frac{1}{n+1} \]

5. **Calculate the probability that at least one event occurs:**
- The probability that at least one event occurs is given by:
\[ p(A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_n) = 1 - p(A_1' \cap A_2' \cap A_3' \cap \ldots \cap A_n') \]
- Substituting the value we found:
\[ = 1 - \frac{1}{n+1} = \frac{n+1 - 1}{n+1} = \frac{n}{n+1} \]

### Final Answer:
The probability that at least one of the events occurs is:
\[ \frac{n}{n+1} \]