Conditional Probability and Independence
4.1 Conditional Probability
Basic Exercises
4.1 The answers will vary. For example, consider an experiment where a balanced die is rolled twice. Let A be the event that 6 occurs on the 1st roll, and let B be the event that 5 occurs on the 2nd roll. Then P (A) = 366 = 16 and P (A|B) = 16. Thus, P (A|B) = P (A).
4.2 a) Fix an arbitrary event E ⊂ Ω. Since P (E|Ω) and P (E) both compute the probability of event E given the same sample space Ω, then one must have P (E| Ω) = P (E).
b) For all events E ⊂ Ω, P (E | Ω) = P (E∩Ω)P (Ω) = P (E)1 = P (E).
c) By (a), one can view unconditional probabilities as conditional, where the event on which one conditions is the sample space Ω.
4.3 a) 4/52 = 1/13, i.e. a king is selected with probability 1/13.
b) 4/12 = 1/3, i.e. given that a face card is selected, it is a king with probability 1/3.
c) 1/13, i.e. given that a heart is selected, it is a king with probability 1/13.
d) 0, i.e. given that a non-face card is selected, the probability that it is a king is 0.
e) 12/52=3/13, i.e. the probability of selecting a face card is 3/13.
f ) 1, i.e. given that a king is selected, the probability that a face card is selected equals 1.
g) 3/13, i.e. given that a heart is selected, the probability that it is a face card equals 3/13.
h) 8/48=1/6, i.e. given that the card chosen is not a king, the probability that a face card is selected equals to 1/6.
4.4 a) 23, 468
471+1, 470+11, 715+23, 468+24, 476+21, 327+13, 782+15, 647 = 23, 468
112, 356 ≈ 0.2089 b) 23, 468
112, 356− 471 = 23, 468
111, 885 ≈ 0.20975 c) 1, 470 + 11, 715 + 23, 468
111, 885 = 36, 653
111, 885 ≈ 0.3276
d) Approximately 20.89% of all U.S. housing units have exactly four rooms. Among the U.S.
housing units consisting of at least two rooms, 20.975% have exactly four rooms. Among the
U.S. housing units with at least two rooms, 32.76% have two, three or four rooms.
4.5 a) 320
1164 ≈ 0.2749 b) 36
253 ≈ 0.1423
c) Part (a): Approximately 27.49% of all faculty members are Assistant Professors. Part (b):
Approximately 14.23% of the faculty members in their 50s are Assistant Professors.
d) The table describing the conditional distribution of rank by age and the marginal probability distribution of rank is given by:
Age
Rank
< 30 30− 39 40 − 49 50 − 59 60+ P (Ri) Full professor (R1) 682 40252 156348 145253 7593 1164430 Associate professor (R2) 683 170402 125348 25368 1593 1164381 Assistant professor (R3) 5768 163402 34861 25336 933 1164320 Instructor (R4) 686 40217 3486 2534 0 116433
Total 1 1 1 1 1 1
4.6 a) 42/94≈ 0.4468; b) 19/94 ≈ 0.2021; c) 11/19 ≈ 0.5789; d) 11/42 ≈ 0.2619;
e) Approximately 44.68% of players on the New England Patriots roster are rookies. Approx-imately 20.21% of players on the New England Patriots roster weigh more than 300 pounds.
57.89% of rookies on the team weigh more than 300 pounds. Approximately 26.19% of the players, who weigh over 300 pounds, are rookies.
4.7 a) Probability distribution of weight for rookies is given by:
Weight P (Wi)
Under 200 (W1) 9/42≈ 0.2143 200− 300 (W2) 22/42≈ 0.5238
Over 300 (W3) 11/42≈ 0.2619
b) Probability distribution of years of experience for players whose weight is over 300 pounds is given by:
Years of Experience P (Yi) Rookie (Y1) 11/19≈ 0.579
1− 5 (Y2) 7/19≈ 0.368 6− 10 (Y3) 1/19≈ 0.053
10 + (Y4) 0
c) Conditional distribution of weight by years of experience and the marginal distribution of weight are given in the following table:
Years of Experience
Weight
Rookie 1− 5 6 − 10 10+ P (Wi) Under 200 (W1) 9/42 9/41 2/8 1/3 21/94 200− 300 (W2) 22/42 25/41 5/8 2/3 54/94 Over 300 (W3) 11/42 7/41 1/8 0 19/94
Total 1 1 1 1 1
d) Conditional probability distribution of years of experience by weight and the marginal prob-ability distribution of years of experience are given by:
Weight
Years of Experience
Under 200 200− 300 Over 300 P (Yi) Rookie (Y1) 9/21 22/54 11/19 42/94
1− 5 (Y2) 9/21 25/54 7/19 41/94
6− 10 (Y3) 2/21 5/54 1/19 8/94
10 + (Y4) 1/21 2/54 0 3/94
Total 1 1 1 1
4.8 a) 0.529; b) 0.153; c) 0.084; d) 0.084/0.153≈ 0.549; e) 0.084/0.529 ≈ 0.1588;
f ) 52.9% of all U.S. citizens 15 years of age or older live with spouse. 15.3% of all U.S. citizens 15 years of age or older are over 64. 8.4% of all U.S. citizens 15 years of age or older live with spouse and are older than 64. Among the U.S. citizens who are over 64, 54.9% live with spouse.
Among the U.S. citizens who are 15 years of age or older and live with spouse, 15.88% are over 64 years old.
g) Conditional probability distributions of age by living arrangement and the marginal proba-bility distribution of age are given by:
Living Arrangement
Age
Alone With spouse With others P (Ai) 15− 24 (A1) 6/126 17/529 154/345 0.177 25− 44 (A2) 38/126 241/529 122/345 0.401 45− 64 (A3) 35/126 187/529 47/345 0.269 Over 64 (A4) 47/126 84/529 22/345 0.153
Total 1 1 1 1
h) Conditional probability distributions of living arrangement by age and the marginal proba-bility distribution of living arrangement are given by:
Age
Living Arrangement
15− 24 25− 44 45− 64 Over 64 P (Li) Alone (L1) 6/177 38/401 35/269 47/153 0.126 With spouse (L2) 17/177 241/401 187/269 84/153 0.529 With others (L3) 154/177 122/401 47/269 22/153 0.345
Total 1 1 1 1 1
4.9 a) 13× 48 × 44 × 40 (13)4× 44 = 1
4, since if we consider the ordered selection of the four cards, then there are 13 spades to choose from for the 1st card, after that there are 48 cards (whose denomination is different from the denomination of the 1st selected card) to choose from for the 2nd card, then (similarly) there are 44 cards to choose from for the 3rd card and finally there are 40 cards to choose from for the 4th card. On the other hand, the event that four cards selected (taking into account the order) have different denominations consists of (13)4× 44 outcomes, since there are (13)4 ways to fix the four different denominations for the 1st, 2nd, 3rd and 4th cards selected and for every such choice, there are 44 ways to pick the specific cards.
b) 12× 39 × 26 × 13 52× 39 × 26 × 13 = 12
52 = 3
13, since there are 12 face cards in the deck, and with every subsequent selection of the cards, there are 13 cards less to choose from, since the suits must be all different.
4.10 0.141; Let X be the number of times that 3 occurs (in 12 tosses). Then P (X≥ 4 | X ≥ 1) = P ((X ≥ 4) ∩ (X ≥ 1)) 4.11 Assume that boys and girls are equally likely to be born. Then
a) 1/2. b) 1/3.
4.12
#7
2$ × 45
57 ≈ 0.27525; Let us fix the attention on the first 8 tosses. Then there are 57 outcomes where all first seven tosses are different from 6. On the other hand, there are#7
2
$ways to pick which of the first seven tosses result in 4, and there are 45 ways to pick the values for the five tosses that are different from 4 and 6.
4.13 38.776%, since 0.019/0.049 = 19/49≈ 0.38776, by the conditional probability rule.
4.14 a) 14× 24
14, using (b), since conditional probability is a probability measure and, thus, the complementation rule is applicable.
4.15 a) 1/2. b) 1/2.
c) 3/4, since the required conditional probability is equal to the ratio of the area of a trapezoid with vertices at the points (1/2, 0), (1, 0), (1, 1) and (1/2, 1/2) to the area of the rectangle with vertices at the points (1/2, 0), (1, 0), (1, 1) and (1/2, 1).
where we used (a) and the fact that
d) If the events are positively correlated, then, given the occurrence of one of them, the chances that the other event occurs increase. If the events are negatively correlated, then, given the occurrence of one of them, the probability of the other event decreases. If the events are independent, then their probabilities remain unchanged irrespective of the occurrence of one of the events.
iii) For arbitrary mutually exclusive events (Bn, n∈ N ),
PA
where we used that (unconditional) probability is countably additive and events Bn∩ A (where n ∈ N ) are mutually exclusive. Thus, all three Kolmogorov axioms hold for PA(·), implying that PA(·) = P (· | A) is a probability measure.
4.19 Note that P (A∩ B) > 0 implies that P (A) > 0 and P (B | A) > 0. Then P (C| A ∩ B) · P (B | A) =P (C∩ A ∩ B)
P (A∩ B) ·P (A∩ B)
P (A) = P (C∩ A ∩ B)
P (A) = P (B∩ C | A), thus,
P (C| A ∩ B) = P (B∩ C | A) P (B| A) . 4.20 Note that
PA(C| B) = PA(C∩ B)
PA(B) = P (C∩ B | A)
P (B| A) = P (C| A ∩ B),
where the last equality holds by Exercise 4.19. Therefore, conditioning first on the occurrence of A and then on the occurrence of B is the same as conditioning on the occurrence of both A and B at once (i.e. conditioning on A∩ B).
Advanced Exercises
4.21 a) Since for any two events A and B, P (A| B) = P (A∩ B)
P (B) , then, upon multiplying both sides by P (B), one obtains that P (A∩ B) = P (A | B)P (B). Therefore,
P (W ∩ R1) = P (W| R1)P (R1) = 0.12× 0.35 = 0.042, P (W ∩ R2) = P (W| R2)P (R2) = 0.18× 0.60 = 0.108, P (W ∩ R3) = P (W| R3)P (R3) = 0.13× 0.05 = 0.0065,
P (B∩ R1) = P (B| R1)P (R1) = 0.81× 0.35 = 0.2835, P (B∩ R2) = P (B| R2)P (R2) = 0.72× 0.60 = 0.432, P (B∩ R3) = P (B| R3)P (R3) = 0.75× 0.05 = 0.0375, P (O∩ R1) = P (O| R1)P (R1) = 0.07× 0.35 = 0.0245, P (O∩ R2) = P (O| R2)P (R2) = 0.10× 0.60 = 0.06, P (O∩ R3) = P (O| R3)P (R3) = 0.12× 0.05 = 0.006 b) Using (a), we obtain that
P (W ) = P (W∩ R1) + P (W ∩ R2) + P (W ∩ R3) = 0.1565, P (B) = P (B∩ R1) + P (B∩ R2) + P (B∩ R3) = 0.753, P (O) = P (O∩ R1) + P (O∩ R2) + P (O∩ R3) = 0.0905
c) Conditional probability distributions of religion by occupation is given by:
Occupation
Religion
W B O
R1 0.042
0.1565 ≈ 0.268 0.28350.753 ≈ 0.376 0.02450.0905 ≈ 0.271 R2 0.15650.108 ≈ 0.690 0.4320.753 ≈ 0.574 0.09050.06 ≈ 0.663 R3 0.00650.1565 ≈ 0.042 0.03750.753 ≈ 0.050 0.09050.006 ≈ 0.066
Total 1 1 1
d) Since P (R1|W ) = 0.268 < 0.35 = P (R1), the randomly chosen member of the community is less likely to be of Religion 1 if you are told that this person is a white-collar worker than if you are not given that information.
4.22 Let Si (i = 1, 2) denote the event that a randomly selected student passes the exam on the ith try. Then
a) P (pass) = P (S1) + P (S1c∩ S2) = 0.6 + P (S2| S1c)P (S1c) = 0.6 + 0.8(1− 0.6) = 0.92 b) P (S1| pass) = P (S1)
P (pass) = 0.6
0.92 ≈ 0.652