Question

If $X_1, X_2, ..., X_n$ are $n$ independent events such that $P(X_r) = \frac{1}{r+1}$, for $r=1, 2, ..., n$, then the probability that none of the $n$ events occur is

• A. $\frac{1}{n}$
• B. $\frac{1}{n+1}$
• C. $\frac{n}{n+1}$
• D. $\frac{n+1}{n+2}$

Hint:

Telescoping Products in Probability. Products like $\prod_r=1^n \fracrr+1$ simplify beautifully due to cancellation.

Answer 1

The Correct Option is B

We are given $n$ independent events $X_1, X_2, \dots, X_n$ with $P(X_r) = \frac{1}{r+1}$. \[ P(\text{none occur}) = \prod_{r=1}^{n} (1 - P(X_r)) = \prod_{r=1}^{n} \left(1 - \frac{1}{r+1}\right) = \prod_{r=1}^{n} \frac{r}{r+1} \] Telescoping product: \[ \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n}{n+1} = \frac{1}{n+1} \]

Answer 2

Step 1: Understand the problem
We have \(n\) independent events \(X_1, X_2, ..., X_n\) where \(P(X_r) = \frac{1}{r+1}\) for \(r = 1, 2, ..., n\). We need to find the probability that none of these events occur.

Step 2: Probability that an event does not occur
Since the events are independent, the probability that none occur is the product of the probabilities that each event does not occur:
\[ P(\text{none occur}) = \prod_{r=1}^n P(X_r^c) = \prod_{r=1}^n \left(1 - P(X_r)\right) \]
Here, \(P(X_r^c)\) is the complement of event \(X_r\).

Step 3: Calculate the complement probabilities
\[ P(X_r^c) = 1 - \frac{1}{r+1} = \frac{r+1 - 1}{r+1} = \frac{r}{r+1} \]

Step 4: Compute the product
\[ P(\text{none occur}) = \prod_{r=1}^n \frac{r}{r+1} = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{n}{n+1} \]
Notice the telescoping pattern where all intermediate terms cancel out:
\[ = \frac{1 \times 2 \times 3 \times \cdots \times n}{2 \times 3 \times 4 \times \cdots \times (n+1)} = \frac{1}{n+1} \]

Final answer: \(\displaystyle P(\text{none of the } n \text{ events occur}) = \frac{1}{n+1}\)