Middle school golf team that has 10golfers. 2 sixth, 2 seventh, 6 eighth what is the probability of 2 being eighth graders
There are a total of 10 golfers on the team, of which 6 are eighth graders.
The probability of selecting an eighth grader on the first pick is 6/10.
Since one golfer has already been selected, there are now 9 golfers left on the team, of which 5 are eighth graders.
The probability of selecting another eighth grader on the second pick is 5/9.
To find the probability of both events occurring (selecting two eighth grade golfers), we multiply the probabilities together:
6/10 * 5/9 = 30/90
Simplifying the fraction, we get:
30/90 = 1/3
Therefore, the probability of selecting two eighth grade golfers from a team of 10 golfers is 1/3 or approximately 0.33.
The question appears incomplete
Are we picking only 2 ?
Then the bot's answer is correct, but it did not say how many we have to pick
You are correct. The question does not specify if we are picking only 2 golfers or not. If we are picking more than 2 golfers, then the answer would be different since the probability of selecting an eighth grader will change with each pick. However, based on the information provided in the question ("what is the probability of 2 being eighth graders"), it seems like we are only picking 2 golfers.
To find the probability of two golfers being eighth graders, you need to first determine the total number of possible combinations of two golfers out of the ten and then calculate how many of those combinations include two eighth graders.
First, let's calculate the total number of possible combinations of two golfers out of ten. This can be done using the formula for combinations, which is "n choose r," where n is the total number of golfers and r is the number of golfers you are choosing.
In this case, you have ten golfers, so n = 10. And you are choosing two golfers, so r = 2. The formula for combinations is:
C(n, r) = n! / (r!(n-r)!)
Plugging in the values, we have:
C(10, 2) = 10! / (2!(10-2)!)
= 10! / (2!8!)
Calculating the factorial values:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
2! = 2 x 1
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Simplifying the calculation:
C(10, 2) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ((2 x 1) x (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1))
Now, let's calculate how many combinations include two eighth graders. Out of the six eighth graders, you need to choose two, so the number of combinations is:
C(6, 2) = 6! / (2!(6-2)!)
= 6! / (2!4!)
Since we have already calculated the factorials in the previous calculation, we can use the same values here. Simplifying:
C(6, 2) = (6 x 5 x 4 x 3 x 2 x 1) / ((2 x 1) x (4 x 3 x 2 x 1))
Now, to find the probability, you divide the number of combinations with two eighth graders by the total number of possible combinations:
Probability = C(6, 2) / C(10, 2)
Substituting the calculated values:
Probability = [(6 x 5 x 4 x 3 x 2 x 1) / ((2 x 1) x (4 x 3 x 2 x 1))] / [(10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / ((2 x 1) x (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1))]
Simplifying:
Probability = (6 x 5 x 4 x 3 x 2 x 1) / (10 x 9 x 8 x 7)
Now, let's calculate:
Probability = (720) / (5,040)
Probability = 0.14285714285714285
Therefore, the probability of two golfers being eighth graders is approximately 0.1429, or 14.29%.