What is the probability of this happening?
in [Math]
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From: Mike Harrison on 16 Jul 2010 01:42 Hi, I've got a question about probability. I'm sure I managed to do these at school, but that was many years ago! Given an event with a 1 in 125 chance of occurring, what is the probability of that event happening exactly 3 out of 8 times? And also the probability of the event happening *at least* 3 out of 8 times? If anyone could explain how this is calculated, that would be really helpful. Many thanks!
From: Gerry Myerson on 16 Jul 2010 01:56 In article <3b00fdf9-4782-4e57-b35d-902d612be03a(a)e5g2000yqn.googlegroups.com>, Mike Harrison <mjh626(a)yahoo.co.uk> wrote: > Hi, I've got a question about probability. I'm sure I managed to do > these at school, but that was many years ago! > > Given an event with a 1 in 125 chance of occurring, what is the > probability of that event happening exactly 3 out of 8 times? Assuming independence, (8-choose-3) x (124-to-the-5th) / (125-to-the-8th). 8-choose-3 is (8 x 7 x 6) / (1 x 2 x 3) = 56. > And also the probability of the event happening *at least* 3 out of 8 > times? That's a little messier. It's 1 minus (probability of 0 plus probability of 1 plus probability of 2) where those individual probabilities are calculated in a similar way to the probability of 3 done earlier. > If anyone could explain how this is calculated, that would be really > helpful. Many thanks! Pick up an intro probability textbook. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: William Elliot on 16 Jul 2010 02:01 On Thu, 15 Jul 2010, Mike Harrison wrote: > Given an event with a 1 in 125 chance of occurring, what is the > probability of that event happening exactly 3 out of 8 times? > > And also the probability of the event happening *at least* 3 out of 8 > times? > > If anyone could explain how this is calculated, that would be really > helpful. Many thanks! Let p be the probality of an event and q = 1 - p the probality of the negation of the event. Then 1 = 1^n = (p + q)^n = sum(k=0,n) C(n,r) p^k q^(n-k) The probablity of the event happening exactly k times in n tries is p_k = C(n,r) p^k q^(n-k); the probablity of it happening at least k times in n tries is sum(j=k,n) p_j; and the probablity of it happening at most k times in n tries is sum(j=0,k) p_j.
From: porky_pig_jr on 16 Jul 2010 02:08 On Jul 16, 1:42 am, Mike Harrison <mjh...(a)yahoo.co.uk> wrote: > Hi, I've got a question about probability. I'm sure I managed to do > these at school, but that was many years ago! > > Given an event with a 1 in 125 chance of occurring, what is the > probability of that event happening exactly 3 out of 8 times? > > And also the probability of the event happening *at least* 3 out of 8 > times? > > If anyone could explain how this is calculated, that would be really > helpful. Many thanks! This is binomial distribution. First, consider a single trial. Let p = probability of event occured (won). p = 0.008. q = 1 - p = probability of event did not occured (lost). q = 0.992. So in a single trial we win with probability p = 0.008, loose with probability q = 0.992. Now. to win exactly 3 out of 8 times means 3 wins and 5 losses, probability of that is (p^3) (q^5) = (0.008^3)(0.992^5). This 3 out of 5 wins can happen in 8C3 ("8 choose 3") ways. You can look up the formula. 8C3 = (8!) / (3! (8-3)!). So you have to multiply by 8C3 and so the answer is (8C3) (0.008^3) (0.992^5). So, again, this is binomial distribution: winning exactly k out of n trials, with p = probability of winning. Now, to win at least 3 out of 8 means to win exactly 3 out of 8 OR exactly 4 out of 8 blah blah blah or exactly 8 out of 8. All these events are mutually disjoint. That is, you CANNOT win, say, exactly 4 *and* 7 out of 8 in the same trial. It's either 4 or 7. And in case of a finite collection of mutually disjoint events, their total probability is the sum of individual probabilities. That is, P(3 out of 8 OR 4 out of 8 OR ... OR 8 out of 8) = P(3 out of 8) + P(4 out of 8) + ... + P(8 out of 8). Each is binomial distribution you already know how to compute. Done.
From: Mike Harrison on 16 Jul 2010 03:06
On Jul 16, 7:08 am, "porky_pig...(a)my-deja.com" <porky_pig...(a)my- deja.com> wrote: > Now. to win exactly 3 out of 8 times means 3 wins and 5 losses, > probability of that is > (p^3) (q^5) = (0.008^3)(0.992^5). This 3 out of 5 wins can happen in > 8C3 ("8 choose 3") ways. > You can look up the formula. 8C3 = (8!) / (3! (8-3)!). So you have to > multiply by 8C3 and so the answer is > (8C3) (0.008^3) (0.992^5). Thanks. So I make that 2.754e-5 or 1-in-36306, is that right? And the calculation for at least 3 out of 8 is: (8C3) (0.008^3) (0.992^5) + (8C4) (0.008^4) (0.992^4) + (8C5) (0.008^5) (0.992^3) + (8C6) (0.008^6) (0.992^2) + (8C7) (0.008^7) (0.992^1) + (8C8) (0.008^8) (0.992^0) = 2.782e-5 or 1-in-35942. Which seems hardly any more likely than the 3 out of 8 on its own. I suppose that's because the chances of getting more than 3 quickly become extremely small. I hope I've calculated it right anyway. |