1) A test has 10 multiple choice questions, and each question has 5 possible answers. If someone were
to guess on each of the 10 questions, what is the probability they get 5 answers correct? What is the
probability they get more than 5 answers correct? How many answers do we expect a random guesser
to get correct?
2) There is an average of 7 automobile accidents on the campus of Cal Poly Pomona each week. What is
the probability there will be 10 accidents in a particular week? What is the probability of less than 3
accidents in a particular week?
3) For the following random variable, find the mean and the variance.
X P(X)
2 0.2
3
5 0.6
1) A test has 10 multiple choice questions, and each question has 5 possible answers. If someone were
to guess on each of the 10 questions, what is the probability they get 5 answers correct? What is the
probability they get more than 5 answers correct? How many answers do we expect a random guesser
to get correct?
Binomial random variable
Success = correct answer, P(success) = 1/5 = 0.2 = p, 1 – p = 0.8 = q, n = 10, X = # of successes
P(X = 5) = 0.0264
P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0064
“expect” indicates “expected value” or “mean”, and mean for a Binomial random variable is np,
so mean = 10(0.2) = 2
2) There is an average of 7 automobile accidents on the campus of Cal Poly Pomona each week. What is
the probability there will be 10 accidents in a particular week? What is the probability of less than 3
accidents in a particular week?
Success = automobile accident, λ = 7 accidents per week, X = # of successes
P(X = 10) = 0.0710
P(X < 3) = P(X = 2) + P(X = 1) + P(X = 0) = 0.0223 + 0.0064 + 0.0009 = 0.0296
3) For the following random variable, find the mean and the variance.
X P(X)
2 0.2
3 0.2
5 0.6
(notice the correct missing probability has been filled in)
Mean = 2(0.2) + 3(0.2) + 5(0.6) = 0.4 + 0.6 + 3 = 4
Variance = (2 – 4)2(0.2) + (3 – 4)2(0.2) + (5 – 4)2(0.6) = 1.6