Review Problems
1.
(a) P(first is good) = 9/12 = 3/4 = 0.75
(b) G1 = first is good
B1 = first is bad
G2 = second is good
P(G2) = P(G2|G1) + P(G2|B1)
= 9/12 * 8/11 + 3/12 * 9/11
= 0.75
(c) P(both are good) = P(G2|G1)
= 9 / 12 * 8/11
= 6/11
(d) P(both are bad) = P(not G2 | not G1)
= P(first is bad) * P(second is bad | first is bad)
= 3/12 * 2/11 = 1/22
(e) P(one is good and one is bad)
= P(G1 ∩ B2) + P(B1 ∩ G2)
= P(B2|G1). P(G1) + P(G2|B1).P(B1)
= 3/11 * 9/12 + 9/11 * 3/12
= 0.409090909
2.
(a) 0.64
(b) 0.04
(c) 0.04
3.
F1 = makes the first shot
F2 = makes the second shot
(a) P(F1 ∩ F2) = P(F2|F1) * P(F1)
= 0.85 * 0.8 = 0.68
(b) P(not F1 ∩ not F2) = P(not F2|not F1) * P(not F1)
= 0.3 * 0.2 = 0.06
(c) P(F1 ∩ not F2) = P(not F2|F1) * P(F1)
= 0.15 * 0.8 = 0.12
4.
M = person is a man
W = person is a woman
L = voted in the last election
(a) P(W ∩ L) = P(L|W) * P(W)
= 0.68 * 0.54
= 0.3672
(b) P(M ∩ not L) = P(not L|M) * P(M)
= 0.38 * 0.46
= 0.1748
(c) P(M|L) = P(M ∩ L) / P(L)
= 0.62 / (0.68 * 0.54 + 0.62 * 0.46)
= 0.950337216 (seems too high!)
5.
(a) P(first is a boy) = 11/24
(b) P(second is a girl ∩ first is a boy)
= 13/23
6.
(a) P(both aces)
= P(first is ace ∩ second is ace)
= P(second is an ace| first is an ace) * P(first is an ace)
= 3/52 * 4/52 = 0.00443787
(b) P(both are spades) = same as above = 0.00443787
(c) Once the first card is chosen, there are 51 - 12 = 39 cards of other suits.
P(choosing another suit) = 39/52
(d) Once the first card is chosen, there are 51 - 3 = 48 cards of other denominations.
P(choosing another card) = 48/52
7.
(a) 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/64
(b) 1/64 (independent events)
(c) 1/64 (independent events)
8.
(a) 1/2 * 1/2 = 1/4
(b) 1/2 * 1/2 * 1/2 = 1/8
(c) (1/2) ^ 10 = 1/1024
9. Repeat question from one of the earlier sections.
10.
Total pop = 48.8 + 50.4 + 74.5 + 66.6 = 240.3 (million)
(a) P(A) = (50.4 + 66.6) / 240.3 = 0.486891386
P(not A) = 1 - P(A) = 0.513108614
(b) P(B) = (48.8 + 50.4 ) / 240.3 = 0.412817312
P(not B) = 0.587182688
(c) P(A ∩ B) = P(A).P(B) = 0.200997193
(d) P(A ∩ not B) = P(A). P(not B) = 0.301288495
(e) P(A|B) = P(A ∩ B) / P (B) = 0.200997193 / 0.412817312 = 0.486891386
(f) P(B|A) = P(B ∩ A) / P(A) = 0.200997193/0.486891386 = 0.412817312
11.
(a)P(The successful key is the first one tried) = 1/3
(b)P(The successful key is the second one tried) = 1/3
(c)P(The successful key is the third one tried) = 1/3
(d)P(The second key works given that the first one did not) = 1/2
13.
P(neither is faulty)
= P(first is not faulty) * P(second is not faulty|first is not faulty)
= 8/12 * 7 /11
= 14/33
14.
Note : P(B) = 0.6 and P(not A) = 0.2
(a) P(A ∩ B) = 0.8 * 0.6 = 0.48
(b) P(A ∪ B) = 0.8 + 0.6 - 0.48 = 0.92
(c) P(B) = 0.6
(d) P(not A ∩ B)
We have, P(not A) = P(not A ∩ B) + P(not A ∩ not B )
=> P(not A ∩ B) = P(not A) - P(not A ∩ not B ) (the second exp's two events are independent, so it's equal to the product of the probabilities of the events)
= P(not A) - P(notA).P(not B)
= 0.2 - 0.2 * 0.4
= 0.12
15.
(a) 1/52
(b)
A = 1st card is not ace of spades
B = 2nd card is ace of spades
P(A∩B) = P(A). P(B|A)
= (1 - 1/52) * (1/51)
= 1/52
(c) equally
(d) 1/52 - see (c)
16.
M = manual check detects
E = electronic check detects
P(not M ∩ not E) = P(not E| not M) * P(not M)
= 0.20 * 0.30 = 0.06
So, 6 % of defective disks are not detected.
17.
TG = this year is good
NG = next year is good
(a) Business conditions both this year and next will be good.
P = (TG ∩ NG) = P(NG | TG) * P(TG)
= 0.7 * 0.6 = 0.42
(b) Business conditions will be good this year and bad next year.
P = P(TG ∩ not NG) = P(not NG|TG) * P(TG)
= 0.3 * 0.6 = 0.18
(c) Business conditions will be bad both years.
P = P(not TG ∩ not NG) = P(not NG|not TG) * P(not TG)
= 0.6 * 0.4 = 0.24
(d) Business conditions will be good next year.
P(NG) = P(NG|TG)P(TG) + P(NG|not TG )P(not TG ) (Form. 4.1 on page 185)
= 0.7 * 0.6 + 0.4 * 0.4
= 0.58
(e) Given that business conditions are good next year, what is the
conditional probability that they were good this year?
P(TG|NG) = P(TG ∩ NG) / P(NG)
= 0.42 / 0.58
= 0.724137931
18.
J = child receives a blue gene from John
M = child receives a blue gene from Maureen
From the questions, it looks like once a gene is given to a child, it cannot be given again to another from that parent. But is that how it really works?
19.
A = person is above ideal weight
B = person has high BP
P(A∩B) = P(A) + P(B) - P(A ∪ B)
= 0.55 + 0.20 - 0.60
= 0.15
P(A) * P(B)
= 0.55 * 0.20
= 0.11
Since the values are different, they are not independent.
21.
(a) What is the probability that the husband earns less than $75,000?
P = (212 + 36)/500 = 0.496
(b) What is the conditional probability that the wife earns more than
$75,000 given that the husband earns more than this amount?
P = P(w> | h>) = P(h> ∩ w>) / P(h>)
= (54/500) / ((198 + 54)/500) = 0.214285714
(c) What is the conditional probability that the wife earns more than
$75,000 given that the husband earns less than this amount?
P = P(w> | h<) = P(w> ∩ h<) / P(h<)
= (36/500) / (248/500) = 0.14516129
(d) Are the salaries of the wife and husband independent?
P(h> ∩ w>) = 54/500 = 0.108
P(h>) = ((198 + 54)/500) = 0.504
P(w>) = 90/500 = 0.18
P(h>) * P(w>) = 0.09072 != 0.108
Hence, the salaries are not independent.
23.
G = purchases gasoline
O = purchases oil
(a) P(G ∩ O) = P(G) + P(O) - P(G ∪ O)
= 0.86 + 0.08 - 0.9
= 0.04
= 4%
(b)
(i) P(O | G) = P(O ∩ G) / P(G)
= 0.04 / 0.86 = 0.046511628
(ii) P(G | O) = P(O ∩ G) / P(O)
= 0.04 / 0.08 = 0.5
(iii) P(O ∩ G) = 0.04
P(O) * P(G) = 0.86 * 0.08 = 0.0688
Hence, not independent.
24.
(a) P(18-24jazz ∩ 35-44jazz)
= P(18-24jazz) * P(35-44jazz)
= 0.14 * 0.10
= 0.014
(b) P(18-24jazz) + P(35-44jazz)
= 0.24
25.
(a) P(M ∩ not W) + P(W ∩ not M)
= P(M)*P(not W) + P(W)*P(not M)
= 3/100 * 95/100 + 97/100 * 5/100
= 0.077
(b) P(M ∪ W)
= P(M) + P(W) - P (M ∩ W)
= P(M) + P(W) - P(M)*P(W)
= (2 + 3)/100 - 2*3/10000
= 0.0494
(c) P(M) * P(W) = 15 * 19/10000 = 0.0285
26.
No, it's not enough as we don't know the overlaps.
27.
(a) P(A ∪ B)
= P(A) + P(B) - P (A ∩ B)
= P(A) + P(B) - P(B|A)*P(A)
and,
P(B) = P(B|A)P(A) + P(B|not A)P(not A) (Form. 4.1 on page 185)
Combining,
P(A ∪ B)
= P(A) + P(B|A)P(A) + P(B|not A)P(not A) - P(B|A)*P(A)
= P(A) + P(B|not A)P(not A)
= 0.6 + 0.1 * 0.4
= 0.64
(b) Assumption being that they are independent,
P(A ∩ B) = P(A) * P(B)
= 0.06
29.
(a) P(A ∪ B)
= P(A) + P(B) - 0 (mut. excl.)
= 0.5
(b) P(A ∪ B)
= P(A) + P(B) - P(A).P(B)
= 0.2 + 0.3 - 0.06
= 0.44
(c) P(A ∩ B ∩ C)
= P(A).P(B).P(C)
= 0.024
(d) P(A ∩ B ∩ C)
= 0 (mutually excl, so they cannot occur together).
30.
B = has breast cancer
M = has positive mammography
P(B|M) = P(M|B)*P(B) / ( P(M|B)*P(B) + P(M|not B)*P(not B) )
= (0.9 * 0.02) / ((0.9 * 0.02) + (0.1 * 0.98) )
= 0.155172414
31.
Note: Identical does not mean it has to be monozygotic. Did not realize this simple fact at first! Also, same sex != identical
Struggled with this for half an hour, and then got the pointer from here - http://www.math.wustl.edu/~feres/Math450Lect02.pdf
BB, GG, BG -> events that the pair has only boys, only girls and mixed.
M, D -> zygoticness
For a given pair,
P(BB|M) = P(GG|M) = 1/2 (equally likely that the pair has both boys, or both girls)
P(GB|M) = 0 (always same sex)
P(GB|D) = 1/2 (sample space - BB, GG, GB, BG, fulfilling events = 2)
P(GG|D) = 1/4 = P(BB|D)
P(GG) = P(GG|M). P(M) + P(GG|D).P(D) (M = !D)
= 1/2 * P(M) + 1/4 * (1 - P(M))
=> 4P(GG) = 2P(M) + 1 - P(M)
=> P(M) = 4P(GG) - 1
Given P(GG) + P(BB) = 0.64
Also, P(M) = 4P(BB) - 1 (assuming equal number of boys and girls)
2P(M) = 4 (P(GG) + P(BB) ) -2
=> P(M) = 2 * 0.64 - 1
= 0.28
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