a probability problem on summation of random variables The number of coins that Josh spots when walking to work is a Poission random variable with mean $6$. Each coin is equally likely to be a penny, a nickel ($5$ cent), a dime ($10$ cent) or a quarter ($25$ cent). Josh ignores the pennies but picks up the other coins. Find the probability that Josh picks up exactly $25$ cents on this way ? 
 A: The numbers of each type of coin found are i.i.d., and each is Poisson with mean $3/2$.  If $N$, $D$ and $Q$ are the numbers of nickels, dimes, and quarters, respectively, then Josh picks up exactly 25 cents iff $(N,D,Q)$ is in the set $\{(5,0,0),(3,1,0),(1,2,0),(0,0,1)\}$.  Since
$$
{\Bbb P}(N=n,D=d,Q=q)=e^{-9/2} \frac{(\frac32)^{n+d+q}}{n! d! q!},
$$
the probability is
$$
\frac{5241}{1280} e^{-9/2}\approx 0.045486.
$$
A: Hints:


*

*the number of each coin is Poisson distributed with parameter $\frac{6}{4}= 1.5$ 

*Ignoring the pennies you can turn this into "The number of coins that Josh spots when walking to work is a Poission random variable with mean 4.5. Each coin is equally likely to be a nickel (5 cent), a dime (10 cent) or a quarter (25 cent). Josh picks up the coins. Find the probability that Josh picks up exactly 25 cents on this way ?"

*Which possible ways are of making 25 cents from nickels, dimes and quarters? 

*What are the probabilities of those being the coins Josh finds?  

*The sum of those probabilities?
