Originally Posted by

**meymathis** So this is asking to find

$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2)$

So break it down. We try to work this stuff out into independent events so that we can calculate the pieces.

Standard equality for conditional probability:

$\displaystyle P(\sum_{i=1}^{10} X_i = 20 | X_1 < 2) =P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 < 2) /P(X_1 < 2)$

Break numerator into disjoint sets.

$\displaystyle = (P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=1}^{10} X_i = 20 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now that $\displaystyle X_1$ is fixed in the two summands, you can write the expressions as what happens to $\displaystyle X_2,\ldots,X_{10}$ and what happens to $\displaystyle X_1$ separately.

$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20 \textrm{ and } X_1 =0)+P(\sum_{i=2}^{10} X_i = 19 \textrm{ and } X_1 =1))/P(X_1 < 2)$

Now we have "and"'s between independent events.

$\displaystyle = (P(\sum_{i=2}^{10} X_i = 20)P(X_1 =0)+P(\sum_{i=2}^{10} X_i = 19)P(X_1 =1))/P(X_1 < 2)$

Now everything is standard Poisson which you should be able to calculate.