A source in the Internal Revenue Service has stated that historically 90% of federal tax returns filed are free of arithmetic errors. A random sample of 25 returns are selected and checked carefully for arithmetic errors. Assuming independence, find the probability that


a. all 25 returns are free of arithmetic errors.
b. at most 23 returns are free of arithmetic errors.
c. more than 17 are free of arithmetic errors

This problem has the following characteristics:

1. Number of trials is fixed (25) and known.
2. The probability of success is constant (0.9) throughout the trials and known.
3. The tests are independent
4. Each test is Bernoulli (i.e. true/false, 0/1)
These criteria indicate the binomial distribution, with
n=25, p=0.9, q=(1-0.9)=0.1
with
P(X=k)=C(n,k)pkqn-k

a.
Calculate P(X=25)

b.
calculate p=P(X=24)+P(X=25)
Probability for at most 23
=1-(P(X=24)+P(X=25))

c.
calculate
25
∑ P(X=k)
k=18

.072

1.075

To find the probabilities requested, we can use the binomial probability formula, as this problem involves a fixed number of trials (25 returns) with two possible outcomes (free of errors or with errors), and the probability of success (a return without errors) is constant for each trial (90%).

The binomial probability formula is given by:

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k),

where:
- P(X = k) is the probability of getting k successes out of n trials.
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k successes out of n trials.
- p is the probability of success.
- (1 - p) is the probability of failure.
- n is the total number of trials.

Now let's calculate the probabilities:

a. Probability that all 25 returns are free of arithmetic errors:
P(X = 25) = C(25, 25) * (0.9^25) * (0.1^(25 - 25)) = (1) * (0.9^25) * (0.1^0) = 0.9^25.

b. Probability that at most 23 returns are free of arithmetic errors:
P(X ≤ 23) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 23).

To calculate this probability, we need to sum the individual probabilities for each value from 0 to 23. However, since this can be a lengthy process, we can use the complement rule to calculate the probability of 24 or 25 returns being error-free:

P(X ≤ 23) = 1 - P(X ≥ 24).

Using the same formula in reverse, we have:
P(X ≥ 24) = P(X = 24) + P(X = 25).

P(X ≥ 24) = C(25, 24) * (0.9^24) * (0.1^1) + C(25, 25) * (0.9^25) * (0.1^0).

c. Probability that more than 17 returns are free of arithmetic errors:

P(X > 17) = 1 - P(X ≤ 17).

Again, using the same formula and the complement rule:
P(X ≤ 17) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 17).

Calculating this probability requires summing the individual probabilities for each value from 0 to 17.

I hope this explanation helps you understand how to calculate the probabilities in this problem!