Question

In a communication network, 98% of messages are transmitted correctly with no error. If the random variable X denotes the number of messages transmitted with error, then the probability that at most two messages are transmitted with error out of 1000 messages sent, is

• A. \( \frac{2}{e^{980}} \)
• B. \( \frac{220}{e^{20}} \)
• C. \( \frac{221}{e^{20}} \)
• D. \( \frac{2}{e^{106}} \)

Hint:

Use Poisson approximation to Binomial when \(n\) is large, \(p\) is small (\(\lambda=np\)).
Poisson PMF: \(P(X=k) = e^{-\lambda}\lambda^k / k!\).
"At most two" means \(k=0, 1, 2\).

Answer 1

The Correct Option is C

Let \( X \) be the number of messages with error. This follows a binomial distribution \( B(n,p) \).

Given: \( n = 1000 \) (number of messages), and probability of error \( p = 1 - 0.98 = 0.02 \).

Since \( n \) is large and \( p \) is small, we can use the Poisson approximation with \( \lambda = np \).

\( \lambda = 1000 \times 0.02 = 20 \), so \( X \sim \text{Poisson}(20) \). 

The Probability Mass Function (PMF) for a Poisson distribution is:

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

We need to calculate \( P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) \).

First, calculate \( P(X = 0) \):

\[ P(X = 0) = \frac{e^{-20} 20^0}{0!} = e^{-20} \]

Next, calculate \( P(X = 1) \):

\[ P(X = 1) = \frac{e^{-20} 20^1}{1!} = 20e^{-20} \]

Then, calculate \( P(X = 2) \):

\[ P(X = 2) = \frac{e^{-20} 20^2}{2!} = \frac{e^{-20} \cdot 400}{2} = 200e^{-20} \]

Now, sum the probabilities to get \( P(X \le 2) \):

\[ P(X \le 2) = e^{-20}(1 + 20 + 200) = 221e^{-20} = \frac{221}{e^{20}} \]

Final Answer: \[ \boxed{\frac{221}{e^{20}}} \]