Six mice are in an experiment on nutrition and 2 are chosen at random for the control group. What is the probability that the two heaviest mice are chosen for the control group?
PLEASE HELP!! This is a review question for an exam I have in the morning!!
looks like just 1/C(6,2) = 1/15
To find the probability that the two heaviest mice are chosen for the control group, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Let's break it down step by step:
Step 1: Determine the total number of mice in the experiment. We are told that there are six mice in total.
Step 2: Determine the number of mice chosen for the control group. We are told that 2 mice are chosen for the control group.
Step 3: Determine the number of favorable outcomes. In this case, we want to choose the two heaviest mice for the control group. Since there are six mice in total, we have 6 options for the first mouse and 5 options for the second mouse. So, the number of favorable outcomes is 6 * 5 = 30.
Step 4: Determine the total number of possible outcomes. This is the number of ways to choose two mice from a group of six. We can use the combination formula, which is given by:
C(n, r) = n! / (r!(n - r)!)
where n is the total number of items, r is the number of items chosen, and ! denotes factorial.
In this case, n = 6 (total number of mice) and r = 2 (number of mice chosen). Plugging these values into the combination formula:
C(6, 2) = 6! / (2!(6 - 2)!) = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / (2 * 1) = 30 / 2 = 15.
So, the total number of possible outcomes is 15.
Step 5: Calculate the probability. The probability of the two heaviest mice being chosen for the control group is given by:
Probability = number of favorable outcomes / total number of possible outcomes
Probability = 30 / 15 = 2/1 = 2.
Therefore, the probability that the two heaviest mice are chosen for the control group is 2/15.
Good luck with your exam!