Finite Math

Question

Suppose that a 4-digit pin number is to be formed with the digits in {0,1,2,3,4,5,6,7,8,9}. What is the probability that the pin number contains at least 3 odd digits?

 

Thanks for the help!

Answer 1

5/16=31.25%

4/16 + 1/16

= .25 + .0625'

= P(3) + P(4)

P(0)= 1/16

P(1)= 4/16

P(2)= 6/16