# 'Average' number

• Jul 29th 2010, 02:11 AM
calbee
'Average' number
Cycle
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1. in each cycle, I will pick up 5 numbers out of 100 numbers (no. 1 to no. 100) randomly
2. in a black box, it contains 100 balls (marked no. 1 to no. 100). One ball will be drawn randomly

Question: In 'Average', how many draws are needed for the ball being drawn from black box would be one of the numbers I picked in (1) ?

Is the above question solvable? Thanks a lot!
• Jul 29th 2010, 01:38 PM
awkward
geometric distribution
Quote:

Originally Posted by calbee
Cycle
----------
1. in each cycle, I will pick up 5 numbers out of 100 numbers (no. 1 to no. 100) randomly
2. in a black box, it contains 100 balls (marked no. 1 to no. 100). One ball will be drawn randomly

Question: In 'Average', how many draws are needed for the ball being drawn from black box would be one of the numbers I picked in (1) ?

Is the above question solvable? Thanks a lot!

Your chance of success on one draw is 5/100, and the number of draws to the first success has a Geometric distribution, so the expected number of draws required is

$\frac{1}{5/100} = 20$.