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

Yes, you are correct in your calculation. The probability of not getting a zero for the first four times is 9/10 for each trial, and the probability of getting a zero on the fifth trial is 1/10. To find the probability that the first zero occurs as the fifth random digit generated, you need to multiply these probabilities together:

P(First zero at fifth trial) = (9/10)^4 * (1/10) = 0.0006561

So the probability is approximately 0.0006561.

To determine how many digits you would expect to generate in order to observe the first zero, you can use the concept of expected value or the average number of trials needed. In this case, the probability of getting a zero on any individual trial is 1/10.

The expected value can be calculated using the formula:
<𝑛> = 1/p

where p is the probability of success (getting a zero) on any individual trial. In this case, p = 1/10.

<𝑛> = 1 / (1/10) = 10

So, you would expect to have to generate 10 digits in order to observe the first zero.

Yes, you are correct in your calculation for the probability that the first 0 occurs as the fifth random digit generated. The probability of not getting a zero for the first four times is indeed 9/10, and the probability of getting a zero for the fifth time is 1/10. Multiplying these probabilities together gives you the probability of the desired outcome.

Regarding your second question, you are on the right track. The expected value is a measure of the average number of trials required to observe a certain event. In this case, you are interested in the expected number of digits you would have to generate in order to observe the first 0.

To calculate this, you can use the concept of expected value. The probability of the first zero occurring on the n-th trial is given by (9/10)^(n-1) * 1/10. The expected value, denoted as <n>, is then calculated as the sum of n times the corresponding probability for each possible value of n:

<n> = sum_{n=1}^∞ n * P(n)

You can evaluate this summation using the formula for the geometric series. By differentiating both sides of the equation ∑ x^n = 1/(1-x), you can obtain the formula for the sum of n times x^(n-1):

sum_{n=1}^∞ n * x^(n-1) = 1/(1-x)^2

Applying this formula with x = 9/10 and multiplying by 1/10, you can calculate the expected value:

<n> = 1/10 * 1/(1 - 9/10)^2 = 10

Therefore, you would expect to have to generate an average of 10 digits in order to observe the first 0.