IE 111 Fall 2008
Homework #3
Officially Due on Friday 9/19
Material on this homework will be fair game for the exam on Wed 9/17
1.
Learn the concept of conditional probability by Venn diagram
a)
If P(AB) =1, must A=B? Draw a Venn diagram to explain your answer?
Answer:
From the Venn diagram, it satisfies P(AB)=1, but A≠B.
b)
Suppose A and B are mutually exclusive events. Construct a Venn diagram that contains the three
events A, B, and C such that P(AC) =1 and P(BC) =0?
2.
Two cards are drawn at random from a deck. Find the probability that both are aces.
Solution:
(4/52) (3/51)= (1/221)
3.
Samples of skin experiencing desquamation are analyzed for both moisture and melanin content.
The results from 100 skin samples are as follows:
Melanin
content
High
low
Moisture
high
13
7
Content
low
48
32
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View Full DocumentLet A denote the event that a sample has low melanin content, and let B denote the event that
sample has high moisture content. Determine the following probabilities:
a) P(A)
b)P(B)
c)
P(AB)
d) P(BA)
Solution:
a) P(A)= (7+32)/100 = 0.4 b) P(B) = (13+7)/100 =0.2
c)
P(AB) = P(A
∩
B) / P(B) = (7/100)/ (20/100) =0.35
d)
P(BA) = P(A
∩
B)/ P(A) = (7/100)/0.4 = 0.175
4.
An integer is selected at random from {1, 2, …, 100}. Given that the number selected is divisible
by 2, find the probability that it is divisible by 3 or 5.
Solution: Let A_2 = event that the number is divisible by 2
Let A_3 = event that the number is divisible by 3
Let A_5 = event that the number is divisible by 5
P(A_3
∪
A_5  A_2) =
P[(A_3
∪
A_5)
∩
A_2 ] / P(A_2)
=
P [(A_3
∩
A_2)
∪
(A_5
∩
A_2) ] / P(A_2)
= P(A_3
∩
A_2) + P (A_5
∩
A_2) – P(A_3
∩
A_5
∩
A_2) / P (A_2)
P(A_3
∩
A_2) = 16/100, P (A_5
∩
A_2) = 10/100, P(A_3
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 Spring '08
 Rodriguez
 Statistics, Conditional Probability, Probability, Probability theory, Lehigh Valley, S. It, desquamation

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