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 July 12th, 2017, 07:17 AM #1 Newbie   Joined: Jul 2017 From: Istanbul Posts: 5 Thanks: 0 Binary Probability Formula Hi I don't good at math so I need a formula. For example there are 4 cells A-B-C-D and their probabilities %40-%60-%45-%55. I want to calculate which binary is the best, A-B or C-D?
 July 12th, 2017, 04:22 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 Your description is is very unclear. is"-" supposed to be a minus sign or a connector? The % add up to 200 - this doesn't make sense.
 July 12th, 2017, 06:54 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond That depends on what assumptions one makes about the data. If the percentages are the chances that, say, four different shooters - each shooting at their own target - are successful, then the pair (C, D) is (possibly) the 'best binary' as 0.45 * 0.55 = 0.2475, whereas 0.40 * 0.60 = 0.24. Whether or not that's a correct interpretation I'll leave to the OP to decide. Thanks from Karma
 July 14th, 2017, 04:52 AM #4 Newbie   Joined: Jul 2017 From: Istanbul Posts: 5 Thanks: 0 They are separate A B C D and %40 %60 %45 %55 and yes I don't care other datas percentages are the chances. So if we change the question A B C D E F... %40 %60 %45 %55 %62 %43... what we need to do is multiply these percentages? If we compare them A B C and D E F 0.40*0.60*0.45 vs 0.55*0.62*0.43 ?
July 14th, 2017, 01:15 PM   #5
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Quote:
 Originally Posted by Karma They are separate A B C D and %40 %60 %45 %55 and yes I don't care other datas percentages are the chances. So if we change the question A B C D E F... %40 %60 %45 %55 %62 %43... what we need to do is multiply these percentages? If we compare them A B C and D E F 0.40*0.60*0.45 vs 0.55*0.62*0.43 ?
When you say A has 40%, what do you mean? Is A a random variable with 2 states (1 or 0)and 40 % is the probability of 1?

When you say A B C do you mean independent variables and all are 1?

You need to clarify your question!

 July 14th, 2017, 03:55 PM #6 Newbie   Joined: Jul 2017 From: Istanbul Posts: 5 Thanks: 0 I got it. For example we have 10 old pistol and each of them has a stuck problem they can't always get fired. Let's say we know their fire chances like first one has %85 other one %92 chance.... Think simple I don't want details other datas, it can be possible to calculate that we took 10 pistol 5 to 5 and shoot to bottles. Which team will win? Last edited by Karma; July 14th, 2017 at 03:57 PM.
 July 15th, 2017, 01:58 PM #7 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 You can the probabilities using the binomial distribution to get the probability for each pistol that it will be successful for n out of 5, where n ranges from 0 to 5. prob (n hits out of 5) = $\displaystyle \binom{5}{n}p^n(1-p)^{5-n}$ p is the probability that a pistol fires. You can do the calculation for each pistol and compare. Thanks from greg1313 and Karma
 July 16th, 2017, 07:44 AM #8 Newbie   Joined: Jul 2017 From: Istanbul Posts: 5 Thanks: 0 Alright I'll change the question. For example I have one dice, one coin and 4 different balls(one of them blue). I throw them at the same time and want to Blue ball, tails, and number 5. Their probabilities 1/4 1/2 1/6 right? What is the total probability?
 July 16th, 2017, 08:41 AM #9 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 $\dfrac{1}{4} * \dfrac{1}{2} * \dfrac{1}{6} = \dfrac{1}{48} \approx 2.1\%.$ The logic is that there are 4 different possible results with the respect to the ball, 2 different possible results with the respect to the coin, and 6 different possible results with the respect to the die, or 48 equally likely possibilities in all. So any one possibility has a probability of 1/48. The formula to remember is: $\text {A and B are independent events} \iff P(A\ and\ B) = P(A) * P(B).$ Thanks from Karma Last edited by JeffM1; July 16th, 2017 at 08:46 AM.
 July 17th, 2017, 01:28 PM #10 Newbie   Joined: Jul 2017 From: Istanbul Posts: 5 Thanks: 0 Thanks

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