Probability
Independent Events
Grade 12
Question:
<p>For independent events \(A_1, A_2, \ldots, A_n\), \(P(A_i) = \frac{1}{i+1}\), \(i = 1, 2, \ldots, n\). Then, the probability that none of the events will occur is</p>
<p>(a) \(\frac{1}{n+1}\)</p>
<p>(b) \(\frac{n}{n+1}\)</p>
<p>(c) \(\frac{1}{(n+1)!}\)</p>
<p>(d) \(\frac{n-1}{n+1}\)</p>
Step-by-Step Solution
Key Concept: Use independence to multiply complementary probabilities and recognize the telescoping product.
<p>For independent events, $P(\text{none occur}) = \prod_{i=1}^{n} P(A_i^c) = \prod_{i=1}^{n} \left(1 - \frac{1}{i+1}\right) = \prod_{i=1}^{n} \frac{i}{i+1} = \frac{1 \cdot 2 \cdot 3 \cdots n}{2 \cdot 3 \cdot 4 \cdots (n+1)} = \frac{1}{n+1}$</p>
Correct Answer: a