Question

A radar system can detect an enemy plane in one out of ten consecutive scans. The probability that it can detect an enemy plane at least twice in four consecutive scans is:

• A. \( 0.0422 \)
• B. \( 0.0523 \)
• C. \( 0.0535 \)
• D. \( 0.0623 \)

Hint:

Use the complement rule when asked for "at least" probabilities in binomial distribution problems: \( P(X \geq r) = 1 - P(X<r) \).

Answer 1

The Correct Option is B

Let the probability of detection in one scan be: \[ p = \frac{1}{10} = 0.1,
q = 1 - p = 0.9 \] We are given 4 consecutive scans and are asked to find the probability of detecting the plane **at least twice**, i.e., \( P(X \geq 2) \), where \( X \sim \text{Binomial}(n = 4, p = 0.1) \) We calculate: \[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \] Step 1: Compute \( P(X = 0) \) \[ P(X = 0) = \binom{4}{0} (0.1)^0 (0.9)^4 = 1 . 1 . 0.6561 = 0.6561 \] Step 2: Compute \( P(X = 1) \) \[ P(X = 1) = \binom{4}{1} (0.1)^1 (0.9)^3 = 4 . 0.1 . 0.729 = 0.2916 \] Step 3: Now, \[ P(X \geq 2) = 1 - 0.6561 - 0.2916 = 0.0523 \]