HW1Solutions
Solutions to Problem Set #1
1. Let Ω = {a, b, c} be a sample space. Let m(a) = 21 , m(b) = probabilities for all eight subsets of Ω.
1
3
and m(c) = 61 . Find the
The eight subsets are: ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}. m(∅) = 0
1
1
1
m(a) = , m(b) = , m(c) =
2
3
6
2
1
5 m({a, b}) = , m({a, c}) = , m({b, c}) =
6
3
2
m({a, b, c}) = 1
6. A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.)
What is the probability of getting an even number in one throw?
If ω1 , · · · , ω6 represent the outcomes of rolling a 1, · · · , 6 on the die (respectively), then the following conditions must hold:
(1) m(ωi ) ≥ 0, i = 1, ..., 6
(2) m(ω1 ) + · · · m(ω6 ) = 1
(3) m(ω6 ) = 2 · m(ω3 ), etc.
(Remember that the first two conditions must always hold for a distribution function).
In order for all three of these conditions to be satisfied simultaneously, we can take
1 + 2 + 3 + 4 + 5 + 6 = 21 to be the denominator for our probability weights. Then
6
1
,...,m(ω6 ) = 21
. (Check for yourself that this choice of values of m(ωi ) satisfies m(ω1 ) = 21 the three conditions above!)
Therefore, P (Even) = P ({2, 4, 6} =
2
21
+
4
21
+
6
21
=
12
21
=
4
7
= 0.57.
7. Let A and B be events such that P (A ∩ B) = 14 , P (Ac ) = 13 , and P (B = 12 . What is
P (A ∪ B)?
Recall Theorem 4 from class: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). We already know that P (B) = 12 and P (A ∩ B) = 14 , so we just need to find P (A). By Theorem 1 part (5), we know that P (Ac ) = 1 − P (A), which is equivalent to saying that P (A) = 1 − P (Ac ).
Therefore,
1
2
P (A) = 1 − P (Ac ) = 1 − = .
3
3
1
Solutions to Problem Set #1
Putting everything together, we have
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
2 1
1
11
+ = = .
3 2
4
12
10. For a bill to come before the president of the United States, it must be passed by both the
House of Representatives and the
1. Let Ω = {a, b, c} be a sample space. Let m(a) = 21 , m(b) = probabilities for all eight subsets of Ω.
1
3
and m(c) = 61 . Find the
The eight subsets are: ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}. m(∅) = 0
1
1
1
m(a) = , m(b) = , m(c) =
2
3
6
2
1
5 m({a, b}) = , m({a, c}) = , m({b, c}) =
6
3
2
m({a, b, c}) = 1
6. A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.)
What is the probability of getting an even number in one throw?
If ω1 , · · · , ω6 represent the outcomes of rolling a 1, · · · , 6 on the die (respectively), then the following conditions must hold:
(1) m(ωi ) ≥ 0, i = 1, ..., 6
(2) m(ω1 ) + · · · m(ω6 ) = 1
(3) m(ω6 ) = 2 · m(ω3 ), etc.
(Remember that the first two conditions must always hold for a distribution function).
In order for all three of these conditions to be satisfied simultaneously, we can take
1 + 2 + 3 + 4 + 5 + 6 = 21 to be the denominator for our probability weights. Then
6
1
,...,m(ω6 ) = 21
. (Check for yourself that this choice of values of m(ωi ) satisfies m(ω1 ) = 21 the three conditions above!)
Therefore, P (Even) = P ({2, 4, 6} =
2
21
+
4
21
+
6
21
=
12
21
=
4
7
= 0.57.
7. Let A and B be events such that P (A ∩ B) = 14 , P (Ac ) = 13 , and P (B = 12 . What is
P (A ∪ B)?
Recall Theorem 4 from class: P (A ∪ B) = P (A) + P (B) − P (A ∩ B). We already know that P (B) = 12 and P (A ∩ B) = 14 , so we just need to find P (A). By Theorem 1 part (5), we know that P (Ac ) = 1 − P (A), which is equivalent to saying that P (A) = 1 − P (Ac ).
Therefore,
1
2
P (A) = 1 − P (Ac ) = 1 − = .
3
3
1
Solutions to Problem Set #1
Putting everything together, we have
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) =
2 1
1
11
+ = = .
3 2
4
12
10. For a bill to come before the president of the United States, it must be passed by both the
House of Representatives and the