Suppose your probability of winning if your number comes up if $\frac{1}{38}$. If so, you win $\$35$, otherwise you lose a dollar. Use the Poisson distribution to show that if you play 70 times then the approximate probability you will have won more money than you have lost is larger than $\frac{1}{2}$
My thoughts are:
Let's say you win k games over the 70 games. You will win more money than you lose if 35k-(70-k) > 0 or k $\ge 2$. So, P(X $\ge$ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – [$\frac{\lambda^0}{0!}e^{-\lambda} + \frac{\lambda^1}{1!}e^{-\lambda}$] = $1-(1 + \lambda) e^{-\lambda}$
However, I'm not getting a prob > $\frac{1}{2}$ if I use $\lambda = \frac{1}{38}$
Is $\lambda$ supposed to be $\frac{1}{38}$ or $\frac{70}{38}$?
Best Answer
Based on Eupraxis's comment, if we use: $\lambda = \frac{70}{38}$
$P(X \ge 2) = 1-\frac{108}{38}e^{-\frac{70}{38}} = 0.5495$