If `A_1,A_2,....A_n` are n independent events such that `P(A_k)=1/(k+1),K=1,2,3,....,n;` then the probability that none of the n events occur is

Options

A. `(n)/(n+1)`
B. `(1)/(n+1)`
C. `(n)/((n+1)(n+2))`
D. none of these

Answer

To find the probability that none of the n independent events \( A_1, A_2, \ldots, A_n \) occur, we can follow these steps:

### Step 1: Understand the given probabilities
We know that the probability of each event \( A_k \) is given by:
\[ P(A_k) = \frac{1}{k + 1} \quad \text{for } k = 1, 2, \ldots, n \]

### Step 2: Calculate the probabilities of the complements
The probability that event \( A_k \) does not occur (the complement of \( A_k \)) is:
\[ P(A_k^c) = 1 - P(A_k) = 1 - \frac{1}{k + 1} = \frac{k + 1 - 1}{k + 1} = \frac{k}{k + 1} \]

### Step 3: Write the expression for the probability that none of the events occur
Since the events are independent, the probability that none of the events \( A_1, A_2, \ldots, A_n \) occur is the product of their complements:
\[ P(A_1^c \cap A_2^c \cap \ldots \cap A_n^c) = P(A_1^c) \cdot P(A_2^c) \cdots P(A_n^c) \]

### Step 4: Substitute the values of the complements
Now we can substitute the values we found for \( P(A_k^c) \):
\[ P(A_1^c) = \frac{1}{2}, \quad P(A_2^c) = \frac{2}{3}, \quad P(A_3^c) = \frac{3}{4}, \ldots, P(A_n^c) = \frac{n}{n + 1} \]
Thus, we have:
\[ P(A_1^c \cap A_2^c \cap \ldots \cap A_n^c) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n}{n + 1} \]

### Step 5: Simplify the product
The product can be simplified as follows:
\[ P(A_1^c \cap A_2^c \cap \ldots \cap A_n^c) = \frac{1 \cdot 2 \cdot 3 \cdots n}{2 \cdot 3 \cdot 4 \cdots (n + 1)} \]
Notice that the numerator is \( n! \) and the denominator can be expressed as:
\[ 2 \cdot 3 \cdots (n + 1) = \frac{(n + 1)!}{1} \]
Thus, we can rewrite the probability as:
\[ P(A_1^c \cap A_2^c \cap \ldots \cap A_n^c) = \frac{n!}{(n + 1)!} = \frac{1}{n + 1} \]

### Final Result
Therefore, the probability that none of the n events occur is:
\[ \boxed{\frac{1}{n + 1}} \]