Answer on Statistics and Probability Question for Aminah
P ( m, X) = m^X *Exp[-m]/X!
m - the expected value of X
P(X=1)= m*Exp[-m] = 0.3
P(X=2)= m^2*Exp[-2m]/2 = 0.2
For example, from the first equation:
m = 0.48
P(X=0) = Exp[-m]= 0.62
P(X=3) = m^3*Exp[-3m]/6 = P(X=1)*P(X=2)/3 = 0.2*0.3/3 = 0.02 &
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