4 ms·

Assuming the chance of each hit is independent, why do you think they wouldn't be additive?

• lostmsu 9y ago
  And potentially add to >100%?
  • chroem- 9y ago
    Yes, that would correspond to multiple missile strikes which is entirely reasonable.
    • wlesieutre 9y ago
      There is no number of missiles you could fire that would guarantee 100% chance of hitting the target (even once, let alone multiple times). 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.
• matt4077 9y ago
  Indendence doesn't play into it. Simple example: flipping two coins doesn't give you a (0.5 + 0.5) = 100% chance of getting heads. 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 Shorter: 1 - P("never six") = 1 - P("first not six, second not six") = 1 - (5/6 * 5/6) = 36/36 - 25/36 = 11/36 The number of possible outcomes you have to list and add together is 2^<number of trials> - 1. For two dices, that's a manageable 2^2 - 1 = 1. For 100 rockets it's 2^100 - 1 = 1.2676506E30 (- 1).
• mnx 9y ago
  They are independent, and that's exactly why they aren't additive. Flipping a coin two times doesn't give you a 100% chance of getting heads. To expand on the intuition a little - they may well hit the same non-target spot multiple times before they hit the target at least once, in fact that is to be expected.
• elsherbini 9y ago
  Intuitively it might make sense to add the probabilities, but it doesn't actually work out that way[0]. The parent used the law of total probability to solve for the probability of hitting at least once because it's simpler: P(hits >= 1) = 1 - P(hits = 0). Solving for the left side of the equation directly gets trickier: 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) Taking that sum where n=100 yields the same as 1 - P(hits = 0) = 1 - (100 chose 0 *0.1^0 *0.99^100) = 1 - 0.99^100 [0] https://en.wikipedia.org/wiki/Binomial_distribution https://en.wikipedia.org/wiki/Binomial_distribution
• deleted 9y ago
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• gregshap 9y ago
  Great explanations, makes perfect sense with a little more thought.
• deleted 9y ago
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