1. The problem statement, all variables and given/known data Suppose that the number of defects on a roll of magnetic recording tape has a Poisson distribution for which the mean λ is either 1.0 or 1.5, and the prior of λ is the following: L(1.0)=0.4 and L(1.5)=0.6 If the roll of tape selected at random is found to have 3 defects, what is the posterior p.f. of λ? 2. Relevant equations p(x|λ) = [e^(-λ)]*(λ^x)/x! L(1.0|X=3) α f(x|λ)L(λ) 3. The attempt at a solution I think this is all I have to do: (1) calculate the probability of X=3 with parameter λ=1 for a poisson distribution and then multiply this by 0.4. Then do the same thing for λ=1.5 and a prior of 0.6. Right? So .4*(1/e)*(1/3!)=0.0245253 and .6*(e^(-1.5))*(1.5^3)/3!=0.0753064 According to the answers from the book, this is wrong. I know the likelihood for the poisson distribution above is e^(-nμ)*μ^(Ʃx)/(x1!*...*xn!) but I don't think I have to calculate this the same way I'd calculate the posterior using a gamma distribution, for example. Any help would be great.