$0 \leq \lim_{n\to \infty}\frac{n!}{(2n)!} \leq \lim_{n\to \infty} \frac{n!}{(n!)^2} = \lim_{k \to \infty, k = n!}\frac{k}{k^2} = \lim_{k \to \infty}\frac{1}{k} = 0.$

2.1-12 Let X be the number of accidents per week in a factory.

Question:

Let X be the number of accidents per week in a factory.  Let the pmf of X be

f(x) = 1/((x+1)(x+2)) = 1/(x+1) - 1/(x+2), x = 0,1,2,...

Find the conditional probability of X >= 4, given that X >= 1.


Answer:

The probability of X>=4 given X >=1 can be written:
Let A be X>=4.
Let B be X>=1.

The probability that A given B, P(A|B) equals...

P(A and B) / P(B).

So we need P(A and B) and P(B).  P(A and B) is the same as P(A), since all X>=4 are also >=1.

Let's make a chart.

x                    f(x)                            
0            1([0]+1) - 1/([0] + 2)    =       .5
1            1([1]+1) - 1/([1] + 2)    =       .16667

2            1([2]+1) - 1/([2] + 2)    =       .08333

3            1([3]+1) - 1/([3] + 2)    =       .05

4            1([4]+1) - 1/([4] + 2)    =       .03333

5            1([5]+1) - 1/([5] + 2)    =       .02381
.
.
.

A is the P(X>=4), which is the complement of P(X<3).  P(X<3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) = .5 +.16667 + .08333 + .05 = .8.  The complement = 1 - 0.8 = 0.2

B is the P(X>=1), which is the same as complement of P(X=0).  1-P(X=0) = 1-0.5 = 0.5


P(A) = 0.2 = 1/5
P(B) = .5 = 1/2

P(A|B) = 1/5 / 1/2 = 0.4