Four students attempt to register online at the same time for an Introductory Statistics class that is full. Two are freshmen and two are sophomores. They are put on a wait list. Prior to the start of the semester, two enrolled students drop the course, so the professor decides to randomly select two of the four wait list students and gives them a seat in the class.What is the probability that both students selected are freshmen? ¼ ⅙ ½ 1/12
Question
Four students attempt to register online at the same time for an Introductory Statistics class that is full. Two are freshmen and two are sophomores. They are put on a wait list. Prior to the start of the semester, two enrolled students drop the course, so the professor decides to randomly select two of the four wait list students and gives them a seat in the class.What is the probability that both students selected are freshmen? ¼ ⅙ ½ 1/12
Solution
The problem can be solved by using the concept of combinations in probability.
First, let's find the total number of ways to select 2 students out of 4. This can be calculated using the combination formula C(n, r) = n! / [(n-r)! * r!], where n is the total number of items, and r is the number of items to choose.
So, the total number of ways to select 2 students out of 4 is C(4, 2) = 4! / [(4-2)! * 2!] = 6.
Next, let's find the number of ways to select 2 freshmen out of 2. This is simply C(2, 2) = 2! / [(2-2)! * 2!] = 1.
Therefore, the probability that both students selected are freshmen is the number of favorable outcomes (selecting 2 freshmen) divided by the total number of outcomes (selecting any 2 students).
So, the probability = 1 / 6 = ⅙.
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