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elkka · #17 04-19-2012, 04:10 AM

Thank you, Professor.

This how I calculate the probabilities. Let $h_{min} = 10*\nu_{min}$ - the number of heads for $c_{min}$ , and let

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.