i have a test tomorrow so i need to know how to do this problem. help me please!!
A IRS auditor randomly selects 3 tax returns from 53 returns of which 13 contain errors. what is the probability that she selects none of those errors?
First one without errors = 40/53
Second one = 39/52
Third one = 38/51
The probability of all events occurring is found by multiplying the probabilities of the individual events.
To find the probability that the IRS auditor selects none of the tax returns with errors, we need to first determine the total number of ways she can select 3 tax returns from the 53 returns. Then, we need to find the number of ways she can select 3 tax returns that do not contain errors.
To calculate this, we will use the concept of combinations. The number of ways to choose k items from a set of n items is given by the formula:
C(n, k) = n! / (k! * (n-k)!)
In our case, the number of ways to select 3 tax returns out of 53 total returns is:
C(53, 3) = 53! / (3! * (53-3)!)
Next, we need to find the number of ways she can select 3 tax returns that do not contain errors. Since there are 13 returns with errors, the remaining 40 returns are error-free. We can apply the combination formula again:
C(40, 3) = 40! / (3! * (40-3)!)
Now, to calculate the probability, we divide the number of successful outcomes (selecting 3 error-free returns) by the total number of possible outcomes (selecting 3 returns from the given set):
P(selecting none of the errors) = C(40, 3) / C(53, 3)
By substituting the values we calculated earlier:
P(selecting none of the errors) = (40! / (3! * (40-3)!)) / (53! / (3! * (53-3)!))
Simplifying the equation, we get the probability that she selects none of the errors in the tax returns.