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July 12th, 2017, 07:17 AM   #1
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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?
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July 12th, 2017, 04:22 PM   #2
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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.
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July 12th, 2017, 06:54 PM   #3
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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.
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July 14th, 2017, 04:52 AM   #4
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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 ?
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July 14th, 2017, 01:15 PM   #5
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Quote:
Originally Posted by Karma View Post
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!
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July 14th, 2017, 03:55 PM   #6
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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.
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July 15th, 2017, 01:58 PM   #7
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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.
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July 16th, 2017, 07:44 AM   #8
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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?
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July 16th, 2017, 08:41 AM   #9
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$\dfrac{1}{4} * \dfrac{1}{2} * \dfrac{1}{6} = \dfrac{1}{48} \approx 2.1\%.$

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:

$\text {A and B are independent events} \iff P(A\ and\ B) = P(A) * P(B).$
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Last edited by JeffM1; July 16th, 2017 at 08:46 AM.
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July 17th, 2017, 01:28 PM   #10
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