eight burglary suspects, two of whom are guilty, are placed in a line up. suppose that a witness is asked to identify the two guilty suspects but is unable to make a positive identification with any degree of confidence. if the witness feels compelled to identify two suspects, what is the probability the witness will pick:
a)two guilty suspects?
b)two innocent subjects, given the first suspect picked is innocent?
To solve this problem, let's first break it down step by step.
Step 1: Determine the total number of ways to choose 2 suspects out of the 8.
The total number of ways to choose 2 suspects out of 8 is given by the combination formula: C(8, 2) = 8! / (2!(8-2)!) = 8! / (2!6!) = (8 * 7) / (2 * 1) = 28.
Now let's move on to the specific scenarios mentioned:
a) Probability of picking two guilty suspects:
Since there are only 2 guilty suspects, let's consider how they can be selected. Since the witness is unable to identify with confidence, they can pick any combination of 2 suspects out of the 8. Therefore, the probability of picking two guilty subjects is equal to the total number of ways to choose 2 suspects (28) divided by the total number of possible outcomes (also 28). So the probability is 28/28 = 1.
b) Probability of picking two innocent suspects, given the first suspect picked is innocent:
If the first suspect picked is innocent, it means we have 7 remaining suspects (1 guilty and 6 innocent) to choose from for the second pick. Therefore, the probability of picking two innocent subjects, given the first suspect picked is innocent, is equal to the number of ways to choose 2 suspects out of the 6 remaining innocent suspects divided by the total number of possible outcomes (28). So the probability is C(6, 2) / 28 = (6! / (2!4!)) / 28 = 15/28.
To summarize:
a) The probability of picking two guilty suspects is 1.
b) The probability of picking two innocent suspects, given the first suspect picked is innocent, is 15/28.