probabiliy geometric ASAP

[howru]

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?

[HallsofIvy]

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.
You are just saying 1/10= 0.1, not '1/10 minus 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.[quote]
You mean that the probability of getting anything other than a zero on EACH of the first four times is 9/10. The probability of not getting a zero after 4 times is (9/10)^4 and, yes, the probability of getting the first zero with the 5th digit is 9^4/10^5.

[quote]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 probality of getting the first 0 on the n th time is 9^(n-1)/10^n. The expected value is the sum of n 9^(n-1)/10^n where the sum is take from 1 to infinity.