# Thread: Discrete Probability Distribution II

1. ## Discrete Probability Distribution II

Could anyone teach me how to solve these?

Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
Q10: (i)0.0803, (ii)0.0498, (iii)0.0032

2. Originally Posted by cloud5
continued from the previous thread... Could anyone teach me how to solve these?

Q9: (i)0.1108, (ii)0.6227, 2, 0.2681
Q10: (i)0.0803, (ii)0.0498, (iii)0.0032
# 9

You are asked to determine the most likely value of X or the mode of this poisson distribution.

$
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$
in which lambda is 2.2

Consider for two cases, if P(X=x+1)>P(X=x), then 2.2>x+1, x<1.2 so x=1

P(X=2)>P(X=1)>P(X=0)

If P(X=x+1)<P(X=x), then 2.2<x+1 , x>1.2 so x=2,3,4,...

P(x=2)>P(X=3)>P(X=4)>...

Do you see that P(X=2) has the highest probability? That implies that the value of X most likely to occur is 2. Calculate its probability.

(10) Adjust the mean to get 1/12.

# 9
$
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$
in which lambda is 2.2
$
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$

How did you find this solution?

4. Originally Posted by cloud5
$
\frac{P(X=x+1)}{P(X=x)}=\frac{\lambda}{x+1}$

How did you find this solution?
It can be proven.
$
P(X=x+1)=\frac{e^{{-\lambda}}{(\lambda)^{x+1}}}{(x+1)!}$
-- 1

$P(X=x)=\frac{{e^{-\lambda}}{\lambda^x}}{x!}$ -- 2

Divide equations 1 and 2, then simplify.