Final answer:
To determine the probability that only 1 out of 3 randomly selected employees has an MBA, calculate the number of ways to choose 1 out of 5 MBAs and 2 out of 15 non-MBAs and divide by the total ways to choose any 3 out of 20 employees. The probability is 35/76.
Step-by-step explanation:
The question asks for the probability that only 1 of the 3 employees selected will have an MBA when simultaneously selecting 3 employees from a total of 20, where 5 have an MBA. This is a probability problem using combinations without replacement.
To solve this, we need to calculate the number of ways to select 1 MBA-holder and 2 non-MBA holders, and then divide that by the number of ways to select any 3 employees from the 20. The number of ways to select 1 MBA-holder is 5 choose 1 (5C1), since there are 5 MBA-holders.
The number of ways to select 2 non-MBA holders is 15 choose 2 (15C2), since there are 15 non-MBA holders among the employees. Thus the numerator of our probability fraction is the product 5C1 * 15C2. And the denominator, the total number of ways to select 3 from 20, is 20 choose 3 (20C3).
The probability is then calculated as:
P(1 MBA) = (5C1 * 15C2) / 20C3
Using the combination formula nCr = n! / (r!*(n-r)!), we find:
P(1 MBA) = (5! / (1!*(5-1)!) * (15! / (2!*(15-2)!))) / (20! / (3!*(20-3)!))
This simplifies to:
P(1 MBA) = (5 / 1 * 105 / 1) / 1140
P(1 MBA) = 525 / 1140
P(1 MBA) = 35 / 76
The probability that only 1 of the 3 employees has an MBA is 35/76.