MGSC 1206.2
SOLUTIONS: Assignment #1
Winter 2008
TOTAL MARKS = 75
Due:
1 pm, Friday, January 18
th
7.3 #22. A single fair die is rolled. Find the probability of the event “getting a number greater than 2”.
Let E = “getting a number greater than 2”
S = {1,2,3,4,5,6} so n(S)=6. The numbers greater than 2 are 3, 4, 5, 6 so n(E) = 4.
P(E) = n(E) / n(S) = 4/6 = 0.67
7.3 #24. A single fair die is rolled. Find the probability of the event “getting any number except 3”.
Let E = “getting any number except 3”
S = {1,2,3,4,5,6} so n(S)=6. The numbers other than 3 are 1, 2, 4, 5, 6 so n(E) = 5.
P(E) = n(E) / n(S) = 5/6 = 0.83
7.3 #44. In 1998, funding for university research in the US totaled $26.343 billion. Support came from
various sources, as shown below.
Source
Amount (in billion $)
Federal government
15.558
State and local government
2.070
Industry
1.896
Academic institutions
4.979
Other
1.840
Find the probabilities that funds for a particular project came from each of the following sources:
(a) federal government,
(b) industry,
(c) academic institutions.
We will assume that the probability that a particular project had funds from a given source is
proportional to the fraction of total funds coming from that source.
a.
P( funds came from federal government ) = 15.558 / 26.343 = 0.591
b.
P( funds came from industry ) = 1.896 / 26.343 = 0.072
c.
P( funds came from academic institutions ) = 4.979 / 26.343 = 0.189
Thus the probabilities of funds coming from federal government, industry, or academic
institutions are 59.1%, 7.2% and 18.9% respectively.
1
2 pts
2 pts
3 pts
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document7.4 #54. The table below shows the probabilities of a person accumulating specific amounts of credit card
charges over a 12month period. Find the probabilities that a person’s total charges during the period are
the following:
(a) $500 or more
(b) less than $1000
(c) $500 to $2999
(d) $3000 or more.
(a) Let E = “total charges are $500 or more”.
P(E)= 1–P(E’) = 1 – [0.31+0.18] = 0.51
OR
P(E) = 0.18 + 0.13 + 0.08 + 0.05 + 0.06 + 0.1 =0.51
The probability that total charges are $500 or more is 51%.
(b) Let E = “total charges are less than $1000”.
P(E) = 0.31 + 0.18 + 0.18 = 0.67
The probability that total charges are less than $1000 is 67%.
(c) Let E = “total charges are $500 to $2999”.
P(E) =
0.18 + 0.13 + 0.08
= 0.39
The probability that total charges are $500 to $2999 is 39%.
(d) Let E = “total charges are $3000 or more”.
P(E) =
0.05 + 0.06 + 0.01 = 0.12
The probability that total charges are $3000 or more is 12%.
7.4 #64. The following data were gathered for 130 adult US workers: 55 were women, 3 women earned
more than $40,000; and 62 men earned $40,000 or less. Find the probabilities that an individual is:
(a) a woman earning $40,000 or less
(b) a man earning more than $40,000
(c) a man or is earning more than $40,000
(d) a woman or is earning $40,000 or less.
$40000 or less
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Jang
 Conditional Probability, Probability

Click to edit the document details