A history class is comprised of
6 female and 9 male students. If the instructor of the class randomly chooses 5 students from the class for an oral exam, what is the probability that 2 female students and 3 male students will be selected? Round your answer to 3 decimal places.
I need help knowing what to do for this problem, so I can do others like it also.
To solve this problem, we need to use the concept of probability and combinations.
First, let's calculate the total number of ways to choose 5 students from the class. We can do this by applying the combination formula:
C(n, r) = n! / ((n - r)! * r!)
where n is the total number of students in the class and r is the number of students being chosen.
In this case, n = 15 (the total number of students in the class) and r = 5 (the number of students being chosen for the oral exam). Plugging these values into the formula, we have:
C(15, 5) = 15! / ((15 - 5)! * 5!) = 3003
There are 3003 different ways to select 5 students from the class.
Next, let's calculate the number of ways to choose 2 female students and 3 male students. We can do this by calculating the combinations separately for each gender, and then multiplying them together:
C(6, 2) = 6! / ((6 - 2)! * 2!) = 15
C(9, 3) = 9! / ((9 - 3)! * 3!) = 84
Therefore, there are 15 ways to choose 2 female students from the 6 available, and 84 ways to choose 3 male students from the 9 available.
To find the probability, we divide the number of successful outcomes (choosing 2 female students and 3 male students) by the total number of possible outcomes (choosing any 5 students from the class):
P(2 female, 3 male) = (15 * 84) / 3003
P(2 female, 3 male) ≈ 0.294
So, the probability that 2 female students and 3 male students will be selected for the oral exam is approximately 0.294, when rounded to 3 decimal places.