For a criminal trial, 8 active and 4 alternate jurors are selected. Two of the alternate jurors are male and two are female. During the trial, two of the active jurors are dismissed. The judge decides to randomly select two replacement jurors from the 4 available alternates. What is the probability that both jurors selected are female? ¼ 1/12 ⅙ ½
Question
For a criminal trial, 8 active and 4 alternate jurors are selected. Two of the alternate jurors are male and two are female. During the trial, two of the active jurors are dismissed. The judge decides to randomly select two replacement jurors from the 4 available alternates. What is the probability that both jurors selected are female? ¼ 1/12 ⅙ ½
Solution
To solve this problem, we need to understand the concept of combinations in probability.
Step 1: Identify the total number of possible outcomes. We have 4 alternate jurors and we need to select 2. The number of ways to do this is given by the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.
So, the total number of ways to select 2 jurors from 4 is C(4, 2) = 4! / [2!(4-2)!] = (4321) / [(21)(21)] = 6.
Step 2: Identify the number of favorable outcomes. We have 2 female alternate jurors and we need to select 2. The number of ways to do this is C(2, 2) = 2! / [2!(2-2)!] = (21) / [(21)*(0!)] = 1. (Note: 0! is defined to be 1.)
Step 3: Calculate the probability. The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
So, the probability that both jurors selected are female is 1 / 6 = 1/6.
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