/sci/ - Science & Math

>>12692599
Not quite - you don't even need to introduce a variable like [math] n [/math]. We know that [math] X [/math] is a Poisson random variable with [math] \lambda = 2 [/math], and so using the formula for its pmf you can directly calculate [math] P(X = 0), P(X = 1), [/math] and [math] P(X = 2) [/math]. So, since we know these values, and we know that [eqn]
\sum_{k=0}^{\infty}P(Y=k) = P(Y = 0)+P(Y = 1)+P(Y = 2)+P(Y = 3) = P(X = 0)+P(X = 1)+P(X = 2)+P(Y = 3) = 1, [/eqn]
we can solve for [math] P(Y = 3) [/math] directly.