Question

A radar system has a probability of 0.1 of detecting a certain target during a single scan. Find the probability that the target will be detected (1) atleast two times in four executive scans (2) atleast once in 20 consecutive scans.

Answer 1

Given:

$X_i = \left \{ {{1 \ ; \ if \ target \ is \ detected\ \ \ \ } \atop {0 \ ; \ if \ target \ is \ not \ detected}} \right.$

$p(x) = \left \{ {{0.1 \ ; \ x \ = \ 1} \atop {0.9 \ ; \ x \ = \ 0}} \right.$

To find:

Find the probability that the target will be detected:

(1) at least two times in four consecutive scans

(2) at least once in 20 consecutive scans.

Solution:

Now each Xi is independent trial with p = 0.1

1) For 4 consecutive trials

Y = ∑Xi

P[target will be detected at least two times in 4 consecutive scans]

P[Y ≥ 2] = ⁴C₂ (0.1)² (0.9)² + ⁴C₃ (0.1)³ (0.9)¹ + ⁴C₄ (0.1)⁴ (0.9)⁰

P[Y ≥ 2] = (6 * 0.01 * 0.81) + (4 * 0.001 * 0.9) + (1 * 0.0001 * 1)

P[Y ≥ 2] = 0.0486 + 0.0036 + 0.0001

P[Y ≥ 2] = 0.0523

2) For 20 consecutive trials

Y = ∑Xi

P[target will be detected at least once in 20 consecutive scans]

P[Y ≥ 1] = 1 - P[Y = 0]

P[Y ≥ 1] = 1 - ²⁰C₀ (0.1)⁰ (0.9)²⁰

P[Y ≥ 1] = 1 - (0.9)²⁰

P[Y ≥ 1] = 0.8784