help 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?
To find the probability that the auditor selects none of the returns with errors, we need to determine the total number of favorable outcomes (the auditor selects 3 returns without errors) and divide it by the total number of possible outcomes (the auditor selects any 3 returns).
To calculate this, we'll use combinations. The number of ways to choose 3 returns without errors from the 40 (53 minus 13) returns without errors is given by the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where:
n is the total number of returns without errors (40)
r is the number of returns to be selected without errors (3)
Using this formula, we can calculate the number of ways to choose 3 returns without errors:
C(40, 3) = 40! / (3!(40-3)!)
= 40! / (3!37!)
Simplifying this expression:
= (40 * 39 * 38) / (3 * 2 * 1)
= 684,280
Now, let's calculate the total number of possible outcomes, which is the number of ways to choose any 3 returns from the total of 53 returns:
C(53, 3) = 53! / (3!(53 - 3)!)
= 53! / (3!50!)
Simplifying this expression:
= (53 * 52 * 51) / (3 * 2 * 1)
= 22,957
Finally, we can find the probability that she selects none of the returns with errors by dividing the number of favorable outcomes by the number of possible outcomes:
Probability = Favorable outcomes / Possible outcomes
= 684,280 / 22,957
≈ 0.0299
Therefore, the probability that the auditor selects none of the returns with errors is approximately 0.0299 or 2.99%.
To find the probability that the IRS auditor selects none of the tax returns with errors, we can use the concept of combinations.
First, let's calculate the total number of ways the auditor can select 3 tax returns from the 53 available returns. This can be done using the combination formula:
nCr = n! / (r!(n-r)!)
where n is the total number of returns and r is the number of returns chosen.
In this case, n = 53 (total number of returns) and r = 3 (number of returns chosen). So the total number of ways to select 3 returns is:
53C3 = 53! / (3!(53-3)!)
Next, we need to find the number of ways to select 3 tax returns without any errors. Since there are 13 returns with errors and we want to avoid selecting any of them, we need to consider the remaining returns without errors.
The number of ways to select 3 tax returns without any errors is calculated as:
(53 - 13)C3 = 40C3 = 40! / (3!(40-3)!)
Now, we can calculate the probability of selecting none of the returns with errors:
P(selecting none of the returns with errors) = (number of ways to select 3 returns without errors) / (total number of ways to select 3 returns)
P(selecting none of the returns with errors) = 40C3 / 53C3
To calculate this probability, let's simplify the expression and evaluate it:
40C3 = 40! / (3!(40-3)!)
53C3 = 53! / (3!(53-3)!)
P(selecting none of the returns with errors) = (40! / (3!(40-3)!) ) / ( (53! / (3!(53-3)!) )
Evaluate the above expression to find the probability.