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Right now I'm having a problem with a statistics problem. More specifically with a binomial distribution problem.

The problem says:

As far as I know, binomial distribution formula says:

[itex]p(x=k)=\binom{n}{k}(p)^{k}(1-p)^{n-k}[/itex]

In which:

*n is the number of trials

*k is the number of success

*p is the probability of success

*(1-p) is the probability of failure, sometimes denoted as q in some textbooks

I know two things, since we are talking about eight children then n=8. The number of success is k=3, therefore the number of failures must be n-k= 5. At this point I feel I'm going well. However the problem begins that problem isn't giving me the probability of success (p). I first tried to calculate it by knowing that if 8 children means 100% of all the trials then 3 girls means 37.5% (I found this by rule of three); yet I'm not completely sure if that's the right way of finding the probability of success p when problem isn't giving it to us.

Thanks.

The problem says:

*There is a family composed by 8 children. Calculate the probability that 3 of them are girls*

As far as I know, binomial distribution formula says:

[itex]p(x=k)=\binom{n}{k}(p)^{k}(1-p)^{n-k}[/itex]

In which:

*n is the number of trials

*k is the number of success

*p is the probability of success

*(1-p) is the probability of failure, sometimes denoted as q in some textbooks

I know two things, since we are talking about eight children then n=8. The number of success is k=3, therefore the number of failures must be n-k= 5. At this point I feel I'm going well. However the problem begins that problem isn't giving me the probability of success (p). I first tried to calculate it by knowing that if 8 children means 100% of all the trials then 3 girls means 37.5% (I found this by rule of three); yet I'm not completely sure if that's the right way of finding the probability of success p when problem isn't giving it to us.

Thanks.