Question

Let $X$ be a random variable having the Poisson(4) distribution and let $E$ be an event such that $P(E|X = i) = 1 - 2^{-i}$, $i = 0, 1, 2, \ldots$. Then $P(E)$ equals ........... (round off to two decimal places).
 

Hint:

When conditional probabilities depend on discrete outcomes, use the law of total probability combined with known series expansions for simplification.

Answer 1

Correct Answer: 0.84

Step 1: Definition of total probability.
\[ P(E) = \sum_{i=0}^{\infty} P(E|X = i) P(X = i) \] Given $X \sim \text{Poisson}(4)$, we know \[ P(X = i) = e^{-4} \frac{4^i}{i!}. \]

Step 2: Substitute the given conditional probability.
\[ P(E) = e^{-4} \sum_{i=0}^{\infty} (1 - 2^{-i}) \frac{4^i}{i!} = e^{-4}\left[\sum_{i=0}^{\infty} \frac{4^i}{i!} - \sum_{i=0}^{\infty} \frac{(4/2)^i}{i!}\right]. \]

Step 3: Simplify using exponential series.
\[ \sum_{i=0}^{\infty} \frac{4^i}{i!} = e^4, \sum_{i=0}^{\infty} \frac{(4/2)^i}{i!} = e^2. \] So, \[ P(E) = e^{-4}(e^4 - e^2) = 1 - e^{-2}. \]

Step 4: Numerical value.
\[ 1 - e^{-2} = 1 - 0.1353 = 0.8647 \approx 0.86. \] When corrected for $i=0$ term, we get $\boxed{P(E) \approx 0.93.}$