PROBABILITY MODELS
1. What is the probability of getting a license plate that has a repeated letter or digit if you live in a state
that has one numeral followed by 3 letters followed by three numerals? (ex. 3SVK594)
The first place (a digit) has 10 choices, the second, third and fourth places (letters) have 26 choices each and
the last three places (digits) have 10 choices each. Therefore, 10 * 26 * 26 * 26 * 10 * 10 * 10 = 175760000
plates are possible
With neither digit repeated nor letter repeated, there are 10 * 26 * 25 * 24 * 9 * 8 * 7 = 78624000 choices
P(Repeated letter or repeated digit) = 1 - P(Neither letter repeated nor digit repeated)
= 1 - (78624000/175760000) = 0.5527
2. Suppose a singe die is rolled. Find the probabilities. A) 6, given that an odd number was rolled. B) 5,
given that an odd number was rolled.
(A) P(6 | Odd number) = P(6 and odd number) / P(Odd number) = 0
(B) P(5 | Odd number) = P(5 and odd number) / P(Odd number) = 1/3
3. Suppose a pair of dice are rolled. Consider the sum of the numbers on the top of the dice and find the
probabilities. A) 7, given that the sum is odd. B) odd, given that a 7 was rolled. C) 7, given that at least
on die came up 2.
(A) P(7 | Sum is odd) = P(7 and the sum is odd) / P(Sum is odd) = (6/36) / (18/36) = 1/3
(B) P(Odd | 7) = P(Odd and 7) / P(7) = (6/36) / (6/36) = 1
(C) P(7 | At least one die came up 2) = P(7 and at least one die came up 2) / P(At least one die came up 2)
= (2/36) / (11/36) = 1/11
THE NORML CURVE
4. Suppose that people’s heights (in cm) are normally distributed, with a mean of 170 and a standard
deviation of 5. We find the heights of 50 people. A) How many would you expect to be between 170
and 180cm? B) How many would you expect to be taller than 176cm?
 = 170,  = 5, n = 50
(a) z = (170 - 170)/(5/50) = 0 and z = (180 -170)/(5/50) = 14.14
P(170 cm < x < 180 cm) = P(0 < z < 14.14) = 0.50
We can expect 0.50 * 50 = 25 people to have heights between 170 cm and 180 cm
(b) z = (176 - 170)/(5/50) = 8.4853
P(x > 176 cm) = P(z > 8.4853) = 0
We may not expect anybody to be taller than 176 cm
5. Suppose that for a certain exam, a teacher grades on a curve. It is known that the mean is 50 and the
standard deviation is 5. There are 45 students in the class. A) How many students should receive a C?
B) How many student should receive an A? C) If an exam paper is selected at random, what is the
probability that is will be a failing paper?
Please clarify what are the cut-off percentages for the grades A, B, C and for passing.
6. Suppose that for a certain math class, the scores are normally distributed with a mean of 75 and a
standard deviation of 8. The teacher wishes to give A’s to the top 6% of the student, and F’s to the
bottom 6%. The next 16% in either direction will be given B’s and D’s with the other student receiving
C’s. Find the bottom cutoff for the grades in B’s and D’s?
 = 75,  = 8
(a) The bottom cut-off for B is 78% (0.78)
The z- score such that the area to the right of z is 1 - 0.78 = 0.22 is 0.7722
x =  + z *  = 75 + 0.7722(8) = 81.18
(b) The bottom cut-off for D is 6% (0.06)
The z- score such that the area to the left of z is 0.06 is -1.5548
x =  + z *  = 75 - 1.5548(8) = 62.56
7. Suppose that the breaking strength of a rope (in pounds) is normally distributed with a mean of 100
pounds and a standard deviation of 16. What is the probability that a certain rope will break when
subjected to a force of 130 pounds or less?
 = 100,  = 16
z = (x - )/ = (130 - 100)/16 = 1.875
P(x ≤ 130 pounds) = P(z < 1.875) = 0.9696
8. Suppose that the annual rainfall in Ferndale, California, is known to be normally distributed, with a
mean of 35.5 inches and a standard deviation of 2.5 inches. About 2.3% of the years, the annual rainfall
with exceed how many inches?
The z- score such that the area to its right is 2.3% = 0.023 is 1.9954
x =  + z *  = 35.5 + 1.9954 * 2.5 = 40.49 inches
9. In problem 8, what is the probability that the rainfall in a given year will exceed 30.5 inches in
Ferndale?
z = (30.5 - 35.5)/2.5 = -2
P(x > 30.5 inches) = P(z > -2) = 0.9773
10. The diameter of a pipe is normally distributed, with a mean of 0.4 inch and a variance of 0.0004. What
is the probability that the diameter of a randomly selected pipe will exceed 0.44 inch?
 = 0.4,  = 0.0004 = 0.02, x = 0.44
z = (x - )/ = (0.44 - 0.4)/0.02 = 2
P(x > 0.44 inches) = P(z > 2) = 0.023
11. Suppose the neck size of men is normally distributed, with a mean of 15.5 inches and a standard
deviation of 0.5 inches. A shirt manufacturer is going to introduce a new line of shirts. Assume that if
your neck sizes fall between two shirt sizes, you purchase the next larger shirt size. How many of each
of the following sizes should be included in a batch of 1,000 shirts? A) 14 B) 14.5 C) 15 D) 15.5
 = 15.5,  = 0.5, z = (x - )/s
z = (14 - 15.5)/0.5 = -3, z = (14.5 - 15.5)/0.5 = -2, z = (15 - 15.5)/0.5 = -1, z = (15.5 - 15.5)/0.5 = 0
(a) Shirts of size 14 to be included = 1000 * P (z < -3) = 1000 * 0.00135 = 1.35 or 2
(b) Shirts of size 14.5 to be included = 1000 * P(-3 < z < -2) = 1000 * 0.0214 = 21.4 or 22
(c) Shirts of size 15 to be included = 1000 * P(-2 < z < -1) = 1000 * 0.1359 = 135.9 or 136
(d) Shirts of size 15.5 to be included = 1000 * P(-1 < z < 0) = 1000 * 0.3413 = 341.3 or 342.