In the September 4, 2006 issue of Space News, Thomas Christie, the former chief weapons tester at the Pentagon, was quoted as saying the Ground Based Mid-course Defense System "likely would have less than a 20% chance of shooting down an incoming missile from North Korea". When asked his take on the system's effectiveness, the president of the Missile Defense Advocacy Alliance, Mr. Riki Ellison, responded "that could be from firing just one missile, and, there are about 9 or 10 interceptors that could take multiple shots at the incoming target, thereby increasing the chances of a hit."

If one interceptor has a 20% chance of hitting a target, what exactly are the chances of a successful defense against an incoming missile with 10 interceptors?

Basic probability theory can help answer that question. Let's start simple. Suppose an antiballistic missile (ABM) interceptor has exactly a 20% chance of hitting an incoming missile. That means it has an 80% chance of missing. Not good.

Suppose two interceptors are shot, and for simplicity, let's assume that each interceptor launched is an independent event (the action of one interceptor does not effect the other). In this case, there are several possible situations that can occur. The first ABM could hit the target, and the second could miss. Likewise, the first ABM could miss and the second one could hit. Or, they both could hit the target, or, both miss it.

Let's abbreviate these four possibilities {HM, MH, HH, MM} using H for a hit and M for a miss. This list of all the possible outcomes for two interceptors is called the *sample space*. An *element* of the sample space is called an *event*. Now assuming each interceptor is independent, and one interceptor does not effect the other, the probability of any compound event of two interceptors can be computed by multiplying the probability of each of the single events.

For example, if both ABMs hit the target, we can multiply the probability of each one hitting to get the probability of the *compound event* of them both hitting:

P(HH) = P(H)P(H)

= (0.20)(0.20)

= 0.04

We have a 4% chance of BOTH missiles hitting the target.

The probability that both *miss* the target is computed as

P(MM) = P(M)P(M)

= (0.80)(0.80)

= 0.64

giving a 64% chance of both ABM's missing the target.

Now, what if one hits, and the other misses:

P(HM) = P(H)P(M)

= (0.20)(0.80)

= 0.16

or a 16% chance of one hitting and one missing. It's the same computation and the same result for the first ABM missing and the second one hitting:

P(MH) = (0.80)(0.20)

= 0.16

Now it is a basic tenent of probability that the sum of all probabilities of events in the sample space must add to 1. We can confirm this by noting that

0.04 + 0.64 + 0.16 + 0.16 = 1

Let's now ask what is the probability that AT LEAST one hit will occur? There are three possibilites in the sample space where at least one hit can occur: HM or MH or HH. To find the probability of at least one hit occuring, we add the probabilities for each of these compound events. (In these simple cases, OR measn ADD in probability.)

P(at least one hit) = P( HM or MH or HH)

= 0.16 + 0.16 + 0.04

= 0.36

So there is a 36% chance that at least one of the two ABMs will it the target missile. Since all probabilities in a sample space must add to 1, and the probability that both interceptors miss is P(MM) = 0.64, we have that:

P(at least one hit) + P(all miss) = 1

0.36 + 0.64 = 1

Subtracting the P(all miss) from both sides of the equation, we have

P(at least one hit) = 1 – P(all miss)

Since there is only ONE WAY that any number of interceptors can ALL MISS the target, this simple equation gives an easier way to compute the probability of at least one hit for any number of ABMs.

With three ABMs you have a sample set of eight different possible outcomes.

{HMM, MHM, MMH, HHM, HMH, MHH, HHH, MMM}

To computer the probabiltiy that at least one of the three interceptors will hit the target, we have

P(at least one hit) = 1 – P(no hit at all)

or

P(HMM or MHM or MMH or HHM or HMH or MHH or HHH) = 1 – P(MMM)

It's easy to compute the probability that all three interceptors miss:

P(MMM) = (0.80)(0.80)(0.80)

= 0.512

= 51.2%

Then, substituting this into our equation, we get the probability that at least one of the interceptors will hit as

P(at least one hit) = 1 – P(all miss)

= 1 – P(MMM)

= 1 – 0.512

= 0.488

= 48.8%

At least with three ABMs we are getting closer to a 50% chance of at least one of them taking out the target.

So what if we had 10 interceptors. The number of total outcomes increases to 2^10. What is the probability that at least one of the ten interceptors will hit the target?

P(at least one hit) = 1 – P(all miss)

= 1 – P(MMMMMMMMMM)

= 1 – (0.8)^10

= 1 – 0.107374

= 0.892626