Assume a target of 0.1km x 0.1km, but the missile may hit anywhere in an area of 1km x 1km around the target. That means the area of the target is 1/100 the size of the overall area. They'd have to launch at least 50 Scuds to have even a 50/50 chance of hitting the target.
Can someone tell me if that calculation is way off? It sounds like this attack, other than spreading fear and panic, was very unlikely to work.
That airport terminal is quite a bit larger, though, and the size of the blast radius of the explosion will increase the likelihood of doing some damage (significantly so for large blast radiuses. If this warhead carried a nuclear device, a miss by a few hundred meters wouldn’t really be a miss)
(1) I guess proper modeling of the error distribution will make that shape something else than a circle, but I also guess a circle is a much better first approximation than a square.
Finally, that attack may have made economic sense. That scud cost a lot less than the rockets fired at it, and showing the capability of firing at targets that far away will pin down quite a few expensive forces around targets in Saudi Arabia.
But, yes: The Scud is terribly (wonderfully) inaccurate.
Think of it like flipping coins. For a given coin flip, 50% odds of heads and 50% odds of tails.
But if you flip the coin 100000000 times there's still a tiny outside chance of tails coming up every time.
As you flip more coins, the odds of getting a heads on at least one of them will asymptotically approach 100%, but it will never get there.
The easiest way to solve "at least once"-type questions is invariably by subtracting P(!"at least once") = P("zero hits") from 1.
To do it with addition, you'd have to list all possible outcomes completely. Those would, individually, be far lower probabilities. Example:
P("at least one "six" when rolling two dice")
= P("first not six, second six") + P("first six, second not six") + P("first six, second also six")
= (1/6 * 5/6) + (5/6 * 1/6) + (1/6 * 1/6)
= 11/36
1 - P("never six")
= 1 - P("first not six, second not six")
= 1 - (5/6 * 5/6) = 36/36 - 25/36
= 11/36
P(hits >= 1) = 1 - P(hits = 0).
P(hits >= 1) = P(hits = 1) + P(hits = 2) + ... + P(hits = k)
P(hits = 1) = 100 chose 1 * 0.01^1 * 0.99^99
...
p(hits = n) = 100 chose n * 0.01^n *0.99^(100-n)
1 - P(hits = 0)
= 1 - (100 chose 0 *0.1^0 *0.99^100)
= 1 - 0.99^100
> During the mujahideen attack against Jalalabad, between March and June 1989, three firing batteries manned by Afghan crews advised by Soviets fired approximately 438 missiles in defense of the embattled garrison.