Question

A radar system can detect an enemy plane in one out of 10 consecutive scans. The probability that it cannot detect an enemy plane at least two times in four consecutive scans, is:

• A. \( 0.9477 \)
• B. \( 0.9523 \)
• C. \( 0.9037 \)
• D. \( 0.9063 \)

Hint:

For problems involving repeated independent trials, use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k}. \] Summing probabilities for multiple cases helps in computing cumulative probabilities.

Answer 1

The Correct Option is A

Step 1: Define Probabilities of Detection and Failure
Let \( p \) be the probability of detecting an enemy plane in a single scan. Given that detection occurs once in 10 scans, we have: \[ p = \frac{1}{10} = 0.1. \] The probability of failing to detect the plane in a single scan is: \[ q = 1 - p = 1 - 0.1 = 0.9. \] Step 2: Define the Probability of Not Detecting at Least Twice in Four Scans
The possible cases where detection occurs fewer than two times in four scans are:
1. No detection in all four scans.
2. Exactly one detection in four scans.
Case 1: No Detection in Four Scans
The probability of missing detection in all four scans: \[ P(0 \text{ detections}) = q^4 = (0.9)^4. \] \[ = 0.6561. \] Case 2: Exactly One Detection in Four Scans
The probability of detecting exactly once in four scans follows the binomial distribution: \[ P(1 \text{ detection}) = \binom{4}{1} p^1 q^3. \] \[ = 4 \times (0.1)^1 \times (0.9)^3. \] \[ = 4 \times 0.1 \times 0.729 = 4 \times 0.0729 = 0.2916. \] Step 3: Compute the Final Probability
The required probability is the sum of these two cases: \[ P(\text{at most 1 detection}) = P(0 \text{ detections}) + P(1 \text{ detection}). \] \[ = 0.6561 + 0.2916 = 0.9477. \] Thus, the correct answer is \( \mathbf{0.9477} \).