X എന്നത് ഒരു പോയിസ്സോൻ ചരമാണ്. P(X=2)=2/3 P(X=1) ആയാൽ P(X=0)=.........A1/e²B1/eCe−43e^{\frac{-4}{3}}e3−4De43e^{\frac{4}{3}}e34Answer: e−43e^{\frac{-4}{3}}e3−4 Read Explanation: P(X=2)=e−λλ22!P(X=2) = \frac{e^{-λ}λ^2}{2!}P(X=2)=2!e−λλ2P(X=1)=e−λλ1!P(X=1)=\frac{e^{-λ}λ}{1!}P(X=1)=1!e−λλP(X=2)=23P(X=1)P(X=2)= \frac{2}{3}P(X=1)P(X=2)=32P(X=1)e−λλ22!=23e−λλ1!\frac{e^{-λ}λ^2}{2!}= \frac{2}{3}\frac{e^{-λ}λ}{1!}2!e−λλ2=321!e−λλλ2=23\frac{λ}{2}= \frac{2}{3}2λ=32λ=43λ=\frac{4}{3}λ=34P(X=0)=e−λλ00!=e−λP(X=0)= \frac{e^{-λ}λ^0}{0!} = e^{-λ}P(X=0)=0!e−λλ0=e−λ=e−43=e^{\frac{-4}{3}}=e3−4 Read more in App