find the probility that when 4 managers are randomly selected from a group of 20, the 4 oldest are selected
The number of quartets from the 20 is
C(20,4) = 20!/(16!4!) = 4845
The "4 oldest" would be one of those groups of 4.
so prob = 1/4845
To solve this problem, we need to calculate the probability of selecting the four oldest managers out of the total group of 20 managers.
Step 1: Calculate the total number of possible ways to select 4 managers out of 20.
This can be done using the combination formula:
nCr = n! / (r! * (n-r)!)
where n is the total number of managers (20) and r is the number of managers selected (4).
Using this formula, we can calculate the total number of combinations:
20C4 = 20! / (4! * (20-4)!) = 20! / (4! * 16!) = (20 * 19 * 18 * 17) / (4 * 3 * 2 * 1) = 4845
Step 2: Calculate the total number of ways to select the four oldest managers.
Since we want to select the four oldest managers, we can assume that they are already specified. Thus, we only need to select the remaining managers from the remaining 16 managers.
16C0 = 1 (There is only one way to not select any managers from a group of 16).
Step 3: Calculate the probability.
To find the probability, we divide the total number of ways to select the four oldest managers by the total number of possible ways to select 4 managers:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = 16C0 / 20C4 = 1 / 4845
Therefore, the probability that when 4 managers are randomly selected from a group of 20, the 4 oldest managers are selected is 1/4845 or approximately 0.000206%.