SOLUTION: Suppose I select a 3-digit positive number at random. What is the solution to P(number less than 900) ?

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose I select a 3-digit positive number at random. What is the solution to P(number less than 900) ?      Log On


   

Question 509491: Suppose I select a 3-digit positive number at random. What is the solution to P(number less than 900) ?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


There are exactly 900 three digit numbers (100 to 999). 800 of them are less than 900. So you have 800 ways to be successful out of 900 possibilities.

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose I select a 3-digit positive number at random. What is the solution to P(number less than 900)
---
3-digit numbers start with 100 and end with 999
That is 999-100+1 = 900 3-digit numbers.
----
The number of those that are less than 900
start at 100 and end with 899
That is 899-100+1 = 800 numbers.
----
P(number less than 900) = 8/9
================================
Cheers,
Stan H.
==================