Posted by david on Tuesday, January 23, 2007 at 12:33am.
When a computerized generator is used to generate random digits, the proability that any particular digit in the set {0,1,2, . . .,9} is generated on any individual trial is 1/10-0.1. suppose that we are generating digits one at a time and are interested in tracking occurrences of the digit 0.
Determine the probabiliy that the first 0 occurs as the fifth random digit generated?
so the probability of not getting a zero for the first four time is 9/10. then the probability of getting a zero for the fifth time is 1/10. and i just multiply that right.
****how many digits would you expect to have to generate in order to observe the first o.
please help me. am i doing this right?
The probability of generating a zero is 1/10, so you would expect to have to generate 10 digits.
You can calculate this from the defintion of the expectation value with a bit of brute force as follows:
The probability that the n-th random digit is the first zero is:
P(n) = (9/10)^(n-1) * 1/10
The expectation value for n is:
<n> = sum_{n=1}^{infinity} n*P(n)
You can calculate this summation using the formula for the geometric series:
sum_{n=0}^{infinity} x^n = 1/[1-x]
differentiate both sides w.r.t. x:
sum_{n=1}^{infinity} nx^(n-1) =
1/[1-x]^(2)
Note that the lower limit of the summation changes to n = 1 because the n = 0 term is a constant which vanishes when differentiated.
Inserting x = 9/10 in here and multiplying by 1/10 gives:
<n> = 1/10 * 1/[1/10]^2 = 10
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