All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
7 come zero (Posted on 2018-06-28) Difficulty: 3 of 5
Let N be a randomly chosen 5-digit 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 n-digit number , n>1.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution Comment 5 of 5 |
(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 (n-1)/10*(9/10)^(n-2)*(n-1)/9*(8/9)^(n-2)=((n-1)^2)/90*(4/5)^(n-2).



  Posted by Math Man on 2018-07-08 19:34:20
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information