Probability Review Problems for Quiz 2

1.     As reported by the FBI, the age distribution of murder victims between 20 and 59 years old is given below.  Use this information to find the probability that a murder victim is:

a.      between 40 and 44 years old, inclusive

b.     25 years old or older

c.     between 45 and 59 years old

d.     under 30 or over 54

 Age # of Murder Victims 20-24 2916 25-29 2175 30-34 1842 35-39 1581 40-44 1213 45-49 888 50-54 540 55-59 372

Solution:

a.     Pr(between 40 - 44) = 1213/11527 = 0.11

b.     Pr(>=25) = 1-Pr(<25) = 1 – (2916/11527) = 1 – 0.25 = 0.75

c.      Pr(between 45 - 59) = (888+540+372)/11527 = 0.16

d.     Pr(<30 or >54) = Pr(<30) + Pr(>54) =

{(2175+2916)/11527} + {372/11527} = 0.44 + 0.03 = 0.47

2.     Two balanced dice are rolled.  Determine the probability that the sum of the dice is

a.      6

b.     even

Solution:

1st, list the sample space:  (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

36 possible outcomes.

Now, list the event set and calculate the probability for each one.

a.     (1,5) (2,4) (3,3) (4,2) (5,1)

Pr(sum=6) = 5/36

b.     (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6)

# Pr(sum is even) = 18/36

3.     According to the Census Bureau, 51% of US adults are female, 7.1% are divorced and 4.1% are divorced females.  For a US adult selected at random, find the following probabilities:

a.      the person is female or divorced

Solution:  Use the general addition rule.

Pr(female or divorced) =

Pr(female) + Pr(divorced) – Pr(female or divorced)=

0.51 + 0.071 – 0.041 = 0.54

4.     Government data assign a single cause for each death that occurs in the U.S.  The data show that the probability is 0.42 that a randomly chosen death was due to heart disease and 0.24 that it was due to cancer.  What is the probability that a death was due to either heart disease or cancer (assuming that only one of these conditions can be assigned as the cause of death)?  What is the probability that it was due to some other cause?

Solution:

Use the addition rule for the first question since the events are mutually exclusive.

Pr(due to heart or cancer) = Pr(heart) + Pr(cancer) = 0.42+0.24 = 0.66

Use the complementation rule for the second question.

Pr(other cause) = 1 – Pr(heart or cancer) = 1 – 0.66 = 0.34

5.     A production process yields 90% satisfactory items.  The quality events pertaining to successive items are assumed to be independent.  What is the probability of finding at least one satisfactory item when the number of selected items is 2.

Solution:  Multiplication Rule and Addition Rule

To find at least one satisfactory, either the first is satisfactory (S1), the second is satisfactory (S2), or both are satisfactory.

Pr(S1U2 or U1S2 or S1S2) = Pr(S1U2) + Pr(U1S2) + Pr(S1S2)=

Pr(S1)Pr(U2) + Pr(U1)Pr(S2) +Pr(S1)Pr(S2)=

0.9*0.1  +  0.1*0.9  +  0.9*0.9 = 0.09+0.09+0.81 = 0.99