From a group of 12 people, you randomly select 3 of them. Find the probability that at least one of them is among the 3 oldest people in the group
Well, if we assume that the ages of the 12 people are all distinct and random, then the probability of selecting at least one of the three oldest people can be calculated as follows:
First, let's find the probability of not selecting any of the three oldest people. To do this, we need to find the number of ways to choose 3 people out of the remaining 9 (since we excluded the three oldest) and divide it by the total number of ways to choose 3 people out of 12. This can be calculated using combinations:
Number of ways to choose 3 people out of 9: C(9,3) = 84
Number of ways to choose 3 people out of 12: C(12,3) = 220
Therefore, the probability of not selecting any of the three oldest people is 84/220.
Since we want the probability of at least one of them being among the three oldest people, we can subtract the probability of not selecting any of them from 1:
P(at least one of the three oldest) = 1 - P(not selecting any of the three oldest)
P(at least one of the three oldest) = 1 - 84/220
P(at least one of the three oldest) = 136/220
Therefore, the probability that at least one of the three randomly selected people is among the three oldest in the group is 136/220, or 34/55.
To find the probability that at least one of the three randomly selected people is among the three oldest in the group, we need to calculate the favorable outcomes and the total possible outcomes.
Step 1: Calculate the total possible outcomes:
To extract 3 people from a group of 12, we use the combination formula:
Total possible outcomes = C(12, 3)
= 12! / (3! * (12-3)!)
= 12! / (3! * 9!)
= (12 * 11 * 10) / (3 * 2 * 1)
= 220
So, there are a total of 220 possible outcomes.
Step 2: Calculate the favorable outcomes:
To select at least one of the three oldest people, we can have any combination of 1, 2, or 3 of them among the selected 3 people. So, the favorable outcomes would be selecting 1, 2, or 3 people from the three oldest.
Favorable outcomes = C(3, 1) + C(3, 2) + C(3, 3)
= (3! / (1! * 2!)) + (3! / (2! * 1!)) + (3! / (3! * 0!))
= (3 * 2 / 2) + (3 / 1) + (1)
= 3 + 3 + 1
= 7
So, there are a total of 7 favorable outcomes.
Step 3: Calculate the probability:
The probability that at least one of the randomly selected people is among the three oldest is given by:
Probability = Favorable outcomes / Total possible outcomes
= 7 / 220
= 0.0318
Therefore, the probability is approximately 0.0318 or 3.18%.
To find the probability that at least one of the three randomly selected people is among the three oldest people in the group, we can use the concept of complementary probability.
First, let's determine the total number of possible outcomes, which is the number of ways to select any three people from the group of 12. This can be calculated using the combination formula.
The number of ways to choose 3 people from a group of 12 is given by:
C(12, 3) = 12! / (3! * (12-3)!) = 220
Now, let's determine the number of favorable outcomes, which is the number of ways to select at least one of the three oldest people.
The number of ways to select at least one of the three oldest people from the group of 12 is the sum of three scenarios:
- Selecting only the oldest person
- Selecting the second-oldest person but not the oldest person
- Selecting the third-oldest person but neither the oldest nor the second-oldest person
For the first scenario, the oldest person can be chosen in C(3, 1) ways. The remaining two people can be chosen from the remaining 11 people in C(11, 2) ways.
For the second scenario, the second-oldest person can be chosen in C(3, 1) ways. The remaining two people can be chosen from the remaining 10 people (excluding the oldest person) in C(10, 2) ways.
Similarly, for the third scenario, the third-oldest person can be chosen in C(3, 1) ways. The remaining two people can be chosen from the remaining 9 people (excluding the two oldest people) in C(9, 2) ways.
Therefore, the number of favorable outcomes is:
C(3, 1) * C(11, 2) + C(3, 1) * C(10, 2) + C(3, 1) * C(9, 2) = 3 * 55 + 3 * 45 + 3 * 36 = 165 + 135 + 108 = 408.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = 408 / 220 ≈ 0.7
So, the probability that at least one of the three randomly selected people is among the three oldest people in the group is approximately 0.7 or 70%.
Alternatively: It's the complement of the probability that none of them is among that group.
Let E[n] be the indexed event that the [n]th person selected is not one of the three oldest.
1 - P(E1) P(E2|E1) P(E3|E1 and E2)
= 1 - (9/12) (8/11) (7/10)
= 0.61818....
There are 12C3 = 220 ways, group of 3 people from 12
There are 9C3 = 84 ways, none of the three oldest.
220-84 = 136 groups where at least one member is 3 oldest.
136/220 = 0..618