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. (Option unreadable/missing in image)
• 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 \(B(n,p)\). \(n = 1000\) (number of messages). Probability of error \(p = 1 - 0.98 = 0.02\). Since \(n\) is large and \(p\) is small, we can use Poisson approximation with \(\lambda = np\). \(\lambda = 1000 \times 0.02 = 20\). So, \(X \sim Poisson(20)\). The PMF is \(P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}\). We need \(P(X \le 2) = P(X=0) + P(X=1) + P(X=2)\). \(P(X=0) = \frac{e^{-20}20^0}{0!} = e^{-20}\). \(P(X=1) = \frac{e^{-20}20^1}{1!} = 20e^{-20}\). \(P(X=2) = \frac{e^{-20}20^2}{2!} = \frac{e^{-20} \cdot 400}{2} = 200e^{-20}\). \(P(X \le 2) = e^{-20}(1 + 20 + 200) = 221e^{-20} = \frac{221}{e^{20}}\). \[ \boxed{\frac{221}{e^{20}}} \]