# Thread: Probability Distribution

1. ## Probability Distribution

Problem: A person is take a multiple choice exam that has 6 questions. Each question has 5 possible choices. The person decides to randomly guess at each problem. Let x be the number of questions the person will answer correctly. Construct the probability distribution, and answer the following:

Find P(x<=2)
Find P(2<=x<=4)
Find the expected value of x
Find the variance of x

What i know: I know the probability of answering a question correctly is (1/5) and the probability for answering a question incorrectly is (4/5). I also know that each question is independent of the others. To find the probabilities for my distribution I used [(1/5)^n]*[(4/5)^m] where n is the number of correctly answered questions and m is the number of incorrectly answered questions. My probability distribution is:

X P(x)
0 .26
1 .07
2 .02
3 .004
4 0
5 0
6 0

This doesn't seem like it could be correct. Any help would be appreciated.

2. Originally Posted by tfhawk
Problem: A person is take a multiple choice exam that has 6 questions. Each question has 5 possible choices. The person decides to randomly guess at each problem. Let x be the number of questions the person will answer correctly. Construct the probability distribution, and answer the following:

Find P(x<=2)
Find P(2<=x<=4)
Find the expected value of x
Find the variance of x

What i know: I know the probability of answering a question correctly is (1/5) and the probability for answering a question incorrectly is (4/5). I also know that each question is independent of the others. To find the probabilities for my distribution I used [(1/5)^n]*[(4/5)^m] where n is the number of correctly answered questions and m is the number of incorrectly answered questions. My probability distribution is:

X P(x)
0 .26
1 .07
2 .02
3 .004
4 0
5 0
6 0

This doesn't seem like it could be correct. Any help would be appreciated.
It cant be correct because the sum over all outcomes is not 1.

The distribution of the number of correct answers is binomial B(6,1/5)

CB