MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 2
Problems from pp. 53-58
11. Let c and C respectively denote the events that a randomly-selected person smokes
cigarettes and cigars. We know that P (c) = 0.28, P (C) = 0.07, and P (c ∩ C) = 0.05.
(a) P (cc ∩C c ) = 1−P (c∪C) = 1−(P (c)+P (C)−P (c∩C)) = 1−(0.28+0.07−0.05) =
0.7.
(b) P (cc ∩ C) = P (C) − P (c ∩ C) = 0.07 − 0.05 = 0.02.
51. Consult your lecture notes.
Theoretical Problems from pp. 59–60
10. Thinking of E ∪ F ∪ G as (E ∪ F ) ∪ G leads you to:
P (E ∪ F ∪ G) = P (E ∪ F ) + P (G) − P ((E ∪ F ) ∩ G)
= P (E ∪ F ) + P (G) − P (EG ∪ F G).
Now P (E ∪ F ) = P (E) + P (F ) − P (E ∩ F ), and P (EG ∪ F G) = P (EG) + P (F G) −
P (EF G), whence the result.
Problems from pp. 104–115
17.
(a) (212 + 36)/500 = 0.496.
(b) 54/(198 + 54) ≈ 0.214.
(c) 36/(212 + 36) ≈ 0.145.
18. Let Ej denote the event that she passes exam j. Thus, we know that P (E1 ) = 0.9,
P (E2 |E1 ) = 0.8, and P (E3 |E1 E2 ) = 0.7.
(a) P (E1 E2 E3 ) = P (E1 )P (E2 |E1 )P (E3 |E1 E2 ) = 0.9 × 0.8 × 0.7 = 0.504.
(b) We want P {E2c | E1c ∪ E2c ∪ E3c }. This is, thanks to the switching lemma,
P
P E2c E1c ∪ E2c ∪ E3c =
E1c
∪
E2c
∪
E3c E2c
P (E2c )
P (E1c ∪ E2c ∪ E3c )
P (E2c )
=
.
P (E1c ∪ E2c ∪ E3c )
(Why?) Because P (E2c ) = 1−P (E2 ) = 1−P (E2 |E1 )P (E1 ) = 1−0.9×0.8 = 0.28,
we have only one computation left:
P (E1c ∪ E2c ∪ E3c ) = 1 − P (E1 E2 E3 ) = 1 − 0.504 = 0.496.
1
Thus, the answer is
0.28
P E2c E1c ∪ E2c ∪ E3c =
≈ 0.565.
0.496
An Additional Remark (June 16, 2003) The exact phrasing of the problem was
misunderstood. The question asks for P (F2 | E1c ∪ E2c ∪ E3c ), where Fi is the event that
exam i was “failed” (versus “not passed.”). Therefore, the correct answer is the following:
=1
}|
{
z
P F2 E1c ∪ E2c ∪ E3c = P E1c ∪ E2c ∪ E3c F2
P (F2 )
∪ E2c ∪ E3c )
(E1c
P
P (F2 | E1 )P (E1 )
=
1 − P (E1 )P (E2 | E1 )P (E3 | E1 E2 )
P (E2c | E1 )P (E1 )
=
1 − P (E1 )P (E2 | E1 )P (E3 | E1 E2 )
0.2 × 0.9
≈ 0.3629.
=
0.504
Theoretical Problems from pp. 115–119
1. Because A ⊂ A ∪ B, P (A) ≤ P (A ∪ B). Therefore, P (A ∪ B) > 0 also, in which case,
P (AB | A ∪ B) =
P (AB)
P (AB)
≤
,
P (A ∪ B)
P (A)
since we are using P (A ∪ B) ≥ P (A) once more. This is equal to P (AB | A).
4. Let Bi denote the event that the ball is in the ith box. Let Si denote the event that
the search in box i found a ball. We know that: P (Bi ) = pi ; and P (Si | Bi ) = αi . We
are asked to find P (Bj |Sic ).
First, if i = j, then by the switching lemma,
P (Bj | Sjc )
P (Sjc |Bj )P (Bj )
P (Sjc |Bj )P (Bj )
=
.
=
P (Sjc )
P (Sjc |Bj )P (Bj ) + P (Sjc |Bjc )P (Bjc )
Now, P (Sjc |Bj )P (Bj ) = (1 −αj )pj . On the other hand, P (Sjc |Bjc )P (Bjc ) = 1(1 −pj ) =
1 − pj . Thus,
(1 − αj )pj
(1 − αj )pj
P (Bj | Sjc ) =
=
,
(1 − αj )pj + 1 − pj
1 − αj pj
as asserted. Likewise, if i 6= j, then
P (Bj | Sic ) =
P (Sic |Bj )P (Bj )
,
P (Sic |Bj )P (Bj ) + P (Sic |Bjc )P (Bjc )
2
but now, P (Sic | Bj ) = 1. Therefore,
P (Bj | Sic ) =
pj +
P (Sic
pj
.
|Bjc )(1 − pj )
(∗)
On the other hand, P (Sic | Bjc ) needs to be computed. Indeed,
P (Sic | Bjc ) = P (Sic | Bi Bjc )P (Bi | Bjc ) + P (Sic | Bic Bjc )P (Bic | Bjc )
P (Bi Bjc ) P (Bic Bjc )
= (1 − αi )
+
.
P (Bjc )
P (Bjc )
Now, P (Bi Bjc ) = P (Bi ) = pi , and P (Bic Bjc ) = 1−P (Bi ∪Bj ) = 1−[P (Bi )−P (Bj )] =
1 − pi − pj . Therefore,
pi
1 − pi − pj
+
1 − pj
1 − pj
pi
pj
= (1 − αi )
+1−
1 − pj
1 − pj
αi pi
=1−
.
1 − pj
P (Sic | Bjc ) = (1 − αi )
Plug this into (∗) to obtain:
P (Bj | Sic ) =
pj + 1 −
pj
.
=
1 − αi pi
3
pj
αi pi
1−pj
(1 − pj )