I'm currently working on the following exercise:
The number of hits, X, per baseball game, has a Poisson distribution. If the probability of a no-hit game is $\frac13$ , what is the probability of having 2 or more hits in specified game?
If I understood correctly the lambda represents the average number of changes we can expect in a given time (for ie).
In this particular case, I'm thinking of as $\lambda = \frac{(2x)}{3}$
is it correct? Is there a formula to find out such variable without much trouble or thinking? Thanks!
If $X$ has Poisson distribution with parameter $\lambda$ then for all nonnegative integers $k$, $\mathbb P(X=k) = e^{-\lambda}\frac{\lambda^k}{k!}$. It is given that $\mathbb P(X=0)=1/3$ (it is written as $13$, but that cannot possibly be a probability, so I will read it as $1/3$), and hence $$ \mathbb P(X=0) = 1/3 = e^{-\lambda}. $$ Taking the logarithm of both sides yields $-\lambda = \log\frac13$ and hence $\lambda = \log 3$. It follows that \begin{align} \mathbb P(X\geqslant 2) &= 1 - (\mathbb P(X=0)+\mathbb P(X=1))\\ &= 1 - e^{-\lambda}(1 + \lambda)\\ &= 1 - e^{-\log 3}(1+\log 3)\\ &= 1 - \frac13(1+\log 3)\approx 0.3004626. \end{align}