Let N be a randomly chosen 5digit number.
a. What is the probability that N contains at least one zero?
b. What is the probability that N contains at least one 7?
c.
What is the probability that it contains exactly one zero and one 7?
d. Generalize the above questions for a ndigit number , n>1.
(In reply to
solution by Charlie)
I disagree with your answer for c. The probability of exactly one 0 is 0.2916. However, given that there is exactly one 0, there are only 9 possibilities for the remaining 4 digits. Therefore, the probability of exactly one 7 is 4*(1/9)*(8/9)^3=2048/6561. Then, the probability of exactly one 0 and one 7 is (729/2500)*(2048/6561)=512/5625. For n digits, the probability is (n1)/10*(9/10)^(n2)*(n1)/9*(8/9)^(n2)=((n1)^2)/90*(4/5)^(n2).

Posted by Math Man
on 20180708 19:34:20 