4 ms·

The chance of a hit is even lower: 1 - (0.99^50) ~= 40%

by virtualized 9y ago

gregshap 9y ago
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
  [deleted]
- gregshap 9y ago
  Great explanations, makes perfect sense with a little more thought.
- deleted 9y ago
  [deleted]

The chance of a hit is even lower: 1 - (0.99^50) ~= 40%

gregshap 9y ago

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
[deleted]