A random variable [tex]X[/tex] following a Poisson distribution with rate parameter [tex]\lambda[/tex] has PMF
[tex]f_X(x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
We're given that [tex]3\mathbb P(X=1)=\mathbb P(X=2)[/tex], so
[tex]\dfrac{3e^{-\lambda}\lambda^1}{1!}=\dfrac{e^{-\lambda}\lambda^2}{2!}[/tex]
[tex]\implies3\lambda=\dfrac{\lambda^2}2[/tex]
[tex]\implies\lambda^2-6\lambda=\lambda(\lambda-6)=0[/tex]
[tex]\lambda[/tex] must be positive, so we arrive at [tex]\lambda=6[/tex], which means
[tex]\mathbb P(X=4)=\dfrac{e^{-6}6^4}{4!}=\dfrac{54}{e^6}\approx0.1339[/tex]
