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:*

* *

*1 ^{st},
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)*

* *

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*