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qr help
Started by Doctor PJ
can anyone tell me how this is done, its from topscore qr.
what is the probability of getting 6 tails out of 9 tosses of a coin?
I would do probability of getting 6 tails of 6 tosses, (1/2)^6 and multiply that by the next 3 flips getting heads * (1/2)^3
(1/2)^6 *(1/2)^3
i think?
You would use the sucess failure formula.
9c6 * (1/2)^6 * (1/2)^3
NcR * Probability of success raised to the power of the number of successes*Probability of failure raised to the power of the number of failures.
1/6 i think without doing the calculation just multiplying in my head. This is he proper way to do this.
9c6 * (1/2)^6 * (1/2)^3
NcR * Probability of success raised to the power of the number of successes*Probability of failure raised to the power of the number of failures.
1/6 i think without doing the calculation just multiplying in my head. This is he proper way to do this.
Last edited:
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chances of getting 6 tails appearing in 9 flips
84/512 -- 16%
just wanted to add. I might be doing this all wrong and unnecessarily long, but I am not sure how to do any other way. Can someone please explain how that NcR success failure works? Thanks.
this is how I did it
9! / [(9-6)!(6!) = 84 , which means there is 84 possible ways to get 6 tails in 9 flips.
Total outcome is 2^9 = 512
84/512 = 16%
Kluzty got 1/6 and it seemed really fast. Can you elaborate on the success failure thing you talked about?
NcR * Probability of success raised to the power of the number of successes*Probability of failure raised to the power of the number of failures.
Always works when there are only 2 possible outcomes, success or failure.
Always works when there are only 2 possible outcomes, success or failure.
how do you know what the probability is out of...you basically have (1/2)^6 x (1/2)^3 = 1/256, now what???
Then multiply by your NcR or in this case 9c6.
Osims,how did u get 256!how do you know what the probability is out of...you basically have (1/2)^6 x (1/2)^3 = 1/256, now what???
Should be 512!
Osims,how did u get 256!
Should be 512!
yea sorry i calculated it wrong in my head my bad your right its supposed to be 1/512
Then I guess you do 9!/6!(9-6)! = 84
so 84 x (1/512) = 84/512
Osims,how did u get 256!
Should be 512!
Correct it should be 512, I didn't bother to check his numbers.
And so in this case if order would matter,we would use permutation one,right????
But the denominator would be the same??? 512????
I believe that is correct although i have never seen a question in my life that involved this formula and had permutations. They always have combinations.
Perhaps streetwolf has may know a drop more.
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