Originally Posted by

**mr fantastic** Question 3:

If you detect k mistakes on a page then obviously $\displaystyle m \geq k$. So:

$\displaystyle \Pr(D = k) = \Pr(D = k \, | \, M = k) \cdot \Pr(M = k) + $ $\displaystyle \Pr(D = k \, | \, M = k + 1) \cdot \Pr(M = k + 1) + \, .....$

$\displaystyle = ^kC_k \, p^k (1 - p)^{k-k} \cdot \frac{e^{-a} a^k}{k!} +$ $\displaystyle ^{k+1}C_k \, p^k (1 - p)^{k + 1 -k} \cdot \frac{e^{-a} a^{k+1}}{(k+1)!} + \, ....$

I suggest you add a couple more terms, simplify each one and take out the common factor. You're left with an infinite sum which doesn't look too bad to find I think.