15 pt Combinatorial probability A 5 character password has j

(15 pt) Combinatorial probability: A 5 character password has just lower case letters and digits and must have at least one letter.

What is the chance of A=exactly 1 character is a letter?

What is the chance of B=all digits are odd?

What is the chance of AB=exactly 1 character is a letter and all digits are odd?

Are the events A and B independent?

Total possible passwords are:

C(5,1)*26*36^4=130*36^4

BEcause we can choose one of 5 places in C(5,1) ways and then 26 letters to choose from

Other 4 places each have:26+10 =36 choices

1. Let\'s count number of passwords satisfying A

Exactly 1 is a letter. We choose one position and fill it with a letter in C(5,1)*26 ways

Rest of the places are only digits and 4 places so can be filled in 10^4 ways

So chance is:

P(A)=10^4/36^4~0.006

2. Let\'s count number of events satisfying B

One of the characters must be a letter. This is done in C(5,1)*26 ways

For other places we have:26+5 choices as only 5 odd digts

So, C(5,1)*26*31^4 passwords

So chance of B,P(B)=31^4/36^4~0.55

3. Lets count passwords of type A intersection B

Exactly one must be letter which is done in C(5,1)*26

Rest four must be odd digits which is done in 5^4

So P(A intersection B)=5^4/36^4=.000372

P(A)P(B)=0.003273

HEnce A and B are not independent