Download Solutions to Assignment 2 (STA 4442)

Solutions to Assignment 2 (STA 4442)
1041
.
1.3-2. (a) 1456
(b) 392
.
633
.
(c) 649
823
(d) The proportion of women who favor a gun law is greater than the proportion of
men who favor a gun law.
1.3-4. (a) We use H denote “heart”. Then, by multiplication rule, we have
P (two hearts) = P (HH) =
13 12
1
·
= .
52 51
17
(b) Let C denote “club”. Then, HC denotes the event that a heart is on the first
draw, and a club is on the second draw. Thus,
P (HC) =
13 13
13
·
=
.
52 51
204
(c) Note that a heart can be drawn by getting the ace of hearts or one of the other
12 hearts. Let A be the event that a heart on the first draw, an ace on the second
draw. Let B be the event that an ace of hearts on the first draw, a non-heart ace on
the second draw and C be the event that a non-ace heart on the first draw, an ace
on the second draw. Then, A = B ∪ C, and B and C are mutually exclusive. So,
P (A) = P (B) + P (C) =
12 4
1
1 3
·
+
·
= .
52 51 52 51
52
1.3-6. Let A be the event that the randomly selected man died of some heart disease,
and B be the event that at least one parent of a died man had some heart disease.
We want to find the conditional probability P (A|B 0 ). By definition, we find
P (A ∩ B 0 )
P (A) − P (A ∩ B)
=
0
P (B )
1 − P (B)
111
221
− 982
110
55
=
=
.
= 982 334
648
324
1 − 982
P (A|B 0 ) =
1.3-10. (a)
P (A) =
52 51 50 49 48 47
·
·
·
·
·
= 0.74141.
52 52 52 52 52 52
1
(b)
0
P (at least two of the drawn cards match) = P (A ) = 1−P (A) = 1−0.74141 = 0.25859.
1.4-2. (a) Since A and B are independent, we have
P (A ∩ B) = P (A)P (B) = (0.3) · (0.6) = 0.18.
Then, by Theorem 1.2-5, we have
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.3 + 0.6 − 0.18 = 0.72.
(b) Since A and B are mutually exclusive, P (A ∩ B) = P (∅) = 0. So,
P (A|B) =
P (A ∩ B)
0
=
= 0.
P (B)
0.6
1.4-10. (a) Note that A is equivalent to the event that 3 is observed on D1 , and 2 is
observed on D2 . So,
3 3
9
P (A) = · = .
4 4
16
(b) Let B1 be the event that 2 is observed on D2 , 1 is observed on D3 . Let B2 be the
event that 5 is observed on D2 , 1 or 4 is observed on D3 . Then, B = B1 ∪ B2 , and
B1 and B2 are mutually exclusive. So,
P (B) = P (B1 ) + P (B2 ) =
9
3 2 1 3
· + · = .
4 4 4 4
16
(c) Let C1 be the event that 1 is observed on D3 , 0 is observed on D1 . Let C2 be the
event that 4 or 6 is observed on D3 , 0 or 3 is observed on D1 . Then, C = C1 ∪ C2 ,
and C1 and C2 are mutually exclusive. So,
P (C) = P (C1 ) + P (C2 ) =
10
2 1 2 4
· + · = .
4 4 4 4
16
1.4-12. (a)
3 2
1
1
P (HHT HT ) + P (H)P (H)P (T )P (H)P (T ) = (P (H)) (P (T )) =
.
2
2
3
2
(b)
3 2
1
1
P (T HHHT ) = P (T )P (H)P (H)P (H)P (T ) = (P (H)) (P (T )) =
.
2
2
3
2
2
(c)
3 2
1
1
.
P (HT HT H) = P (H)P (T )P (H)P (T )P (H) = (P (H)) (P (T )) =
2
2
(d) To have three heads occurring in the five trials, there are 52 different combinations. For each combination (e.g. HHT HT ), the probability is always ( 21 )3 ( 12 )2 as
calculated in (a), (b) or (c). So the desired probability is
3
2
3 2
3 2
5
1
1
5
1
5! 1
1
=
= 10
.
2
2
3!2! 2
2
2
2
1.4-16. (a) Let A be the event that you win when the sampling is done with replacement. Then,
A = {W, LLW, LLLLW, · · · }
which consists of countable number of outcomes. All these outcomes are mutually
exclusive. So, we have
P (A) = P (W ) + P (LLW ) + P (LLLLW ) + · · ·
2
4
1
1
4
4
1
· +
· + ···
+
=
5
5
5
5
5
∞
2k
X1
4
1
1
5
·
=
= ·
2 = .
4
5
5
5 1−
9
k=0
5
(b) Let B be the event that you win when the sampling is done without replacement.
Then,
B = {W, LLW, LLLLW }.
Thus,
P (B) = P (W ) + P (LLW ) + P (LLLLW )
1 4 3 1 4 3 2 1 1
3
=
+ · · + · · · · = .
5 5 4 3 5 4 3 2 1
5
1.5-2. (a) Note that P (A) = 0.4, P (B) = 0.6, P (G|A) = 0.85, P (G|B) = 0.75. By
the Law of Total Probability, we have
P (G) = P (A ∩ G) + P (B ∩ G)
= P (A)P (G|A) + P (B)P (G|B)
= (0.4)(0.85) + (0.6)(0.75) = 0.79.
3
(b) We want to find P (A|G. By Bayes’s formula, we have
P (A|G) =
P (A)P (G|A)
(0.4)(0.85)
P (A ∩ G)
=
=
= 0.43.
P (G)
P (G)
0.79
1.5-8. Let A be the event that the DVD player needed repair under warranty. We
want to find P (Bi |A), i = 1, 2, 3, 4.
By Bayes’s formula, we have
P (B1 )P (A|B1 )
P (B1 )P (A|B1 ) + P (B2 )P (A|B2 ) + P (B3 )P (A|B3 ) + P (B4 )P (A|B4 )
(0.40)(0.10)
=
(0.40)(0.10) + (0.30)(0.05) + (0.20)(0.03) + (0.10)(0.02)
40
40
=
=
= 0.635.
40 + 15 + 6 + 2
63
P (B1 |A) =
Similarly, we can find that
15
= 0.238;
63
6
= 0.095;
P (B3 |A) =
63
2
P (B4 |A) =
= 0.032.
63
P (B2 |A) =
4