Question

A radar unit tracks a target which may use interference. If the target does not employ interference, then during on surveillance cycle the unit may detect it with probability p0. If it employs interference, then the unit may detect it with probability p1 < p0. The probability that the interference will be used in one cycle is p and does not depend on the way and time the interference was used in other cycles. Find the probability that the target will be detected at least once in n surveillance cycles.

Answer 1

The probability that the target will be detected at least once in n surveillance cycles is given by the following expression:
`1 - (1-p_0+p(p_1-p_0))^(n)`

Explanation:

For each surveillance cycle, there are only two possible outcomes. Either it detects the target, or it does not. Cycles are independent. This means that the binomial probability distribution is used to solve this question:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Probability of detecting a target:

p0 of 1-p (without interference)

p1 of p(with interference). So

`\pi = p_0(1-p) + p_1p = p_0 + p(p_1-p_0)`

Find the probability that the target will be detected at least once in n surveillance cycles.

This is

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = x) = C_(n,x).\pi^(x).(1-\pi)^(n-x)`

`P(X = 0) = C_(n,0).(p_0+p(p_1-p_0))^(0).(1-p_0+p(p_1-p_0))^(n) = (1-p_0+p(p_1-p_0))^(n)`

So

`P(X \geq 1) = 1 - P(X = 0) = 1 - (1-p_0+p(p_1-p_0))^(n)`

The probability that the target will be detected at least once in n surveillance cycles is given by the following expression:
`1 - (1-p_0+p(p_1-p_0))^(n)`