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Old 04-19-2012, 03:10 AM
elkka elkka is offline
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Posts: 57
Default Re: question about probability

Thank you, Professor.

This how I calculate the probabilities. Let h_{min} = 10*\nu_{min} - the number of heads for c_{min}, and let h_i be the number of heads in i-th experiment (out of 1000). Then, as Professor has shown previously,

P(\nu_{min}=0) =P(h_{min}=0) = P(\exists i\in[1,1000]\,\, h_i = 0)=1-P(\forall i\in[1,1000]\, h_i > 0)
=1-P(h_1 > 0)^{1000};

Now, P(h_1>0) = 1-P(h_1 = 0) = 1-0.5^{10} . Therefore,
P(\nu_{min}=0)=1-(1-0.5^{10})^{1000} = 0.623576.

Next,
P(\nu_min=0.1)=P(h_{min}=1) =P(\forall i\in[1,1000] \,\,h_i > 0)-P(\forall i\in[1,1000] \,\,h_i > 1)
= P^{1000}(h_1>0)-P^{1000}(h_1>1) = (1-P(h_1=0))^{1000}-(1-P(h_1=0)-P(h_1=1))^{1000}
=(1-0.5^{10})^{1000}-(1-0.5^{10}-C_{10}^1*0.5^{10})^{1000} =0.3764034.

Next,
P(\nu_min=0.2)=P(h_{min}=2) =P(\forall i\in[1,1000]\,\, h_i > 1)-P(\forall i\in[1,1000] \,\,h_i > 2)
= P^{1000}(h_1>1)-P^{1000}(h_1>2)
= (1-P(h_1=0)-P(h_1=1))^{1000}-(1-P(h_1=0)-P(h_1=1)-P(h_1=2))^{1000}
=(1-0.5^{10}-C_{10}^1*0.5^{10})^{1000}-(1-0.5^{10}-C_{10}^1*0.5^{10}-C_{10}^2*0.5^{10})^{1000} =0.000204.

The rest can be calculated directly too, but they are essenctially 0 for the purpose of calculating the mean.
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