Sheldon Ross Introductory Statistics - Probability Section 4.5

Conditional Probability and Independence

1.
(a) A -> obese
B -> diabetes

P(B|A) = P(A ∩ B) / P(A)
= 0.02/0.3 = 0.067

(b) P(A|B) = P(A ∩ B) / P(B)
= 0.02/0.03 = 0.67

2.
A -> first flip is heads
B -> second flip is heads

P(B|A) = P(A ∩ B) / P(A) = 1/4 / 2/4 = 1/2 = 0.5

3.
Mistake in q? There's no bracket for >25k

5.
Total = 28900
(a) Less than 10 years old = 4200/28900 = 0.14532872
(b) Between 10 and 20 years old = 5100 / 28900 = 0.176470588
(c) Between 20 and 30 years old = 6200/28900 = 0.214532872
(d) Between 30 and 40 years old = 4400 / 28900 = 0.152249135

6.
(a) Between 10 and 20 years old, given that the resident is less than 30 years old
A -> between 10 and 20
B -> less than 30

P(A|B) = P(A ∩ B) / P(B) = (5100/28900) / (15500/28900) = 0.329032258

(b) Between 30 and 40 years old, given that the resident is older than 30

A -> between 30 and 40
B -> older than 30

P(A|B) = P(A ∩ B) / P(B) = (4400/28900)/(13400/28900) = 0.328358209

7.
(a)
A -> plays chess
B -> plays bridge

P(A|B) = P(A ∩ B) / P(B) = (26/120) / (56/120) = 0.464285714

(b)
A -> plays bridge
B -> plays chess

P(A|B) = P(A ∩ B) / P(B) = (26/120)/(40/120) = 0.65

Most problems after this are copies with the situations changed. Skipping till I find something interesting.

27.
(a) S = {(U, D), (D, U), (U, U), (D,D)} (move up, then down, or the reverse, over two days)

P(reach original after 2 days) = 2/4 = 1/2

(b) S = {(u,u,u), (u,u,d), (u,d,d), (d,d,d), (d,d,u), (d, u, u), (d,u,d), (u,d,u)}
Risen by 1 after 3 = {(u,u,d), (d, u, u), (u,d,u)}
P(rise by 1 after 3 days) = 3/8

(c) P (went up on first | risen by 1 after 3 days) = P (went up on first ∩ risen by 1 after 3 days) / P (risen by 1 after 3 days)
= (2/8) / (3/8) = 2/3

28.
(a) Independent
(b) Ind
(c) Ind

33.
A = ace
B = spade

P(A) = 4/52
P(B) = 13/52

P(A).P(B) = 1/52

P(A ∩ B) = P(A|B). P(B) = (1/13).(13/52) = 1/52

The probabilities are the same - hence, they are independent.

34.
A = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}
P(A) = 6/36 = 1/6
------------------------------------
B(first die on 1) = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}
P(B) = 6/36 = 1/6

P(A∩B) = 1/36

P(A).P(B) = P(A∩B). Hence A is independent of the fact that the first die lands on 1.
------------------------------------
C(first die on 2) = {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}
P(C) = 1/6

P(A∩C) = 1/36

Independent
------------------------------------
D(first die on 4) = {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}
P(D) = 1/6

P(A∩D) = 1/36

Independent
------------------------------------
E(first die on 5) = {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}
Same. Ind

------------------------------------
F(first die on 6) = {(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}
Same.

Actually, because the sum is constant in all these cases, there is only one possibility for the second die to make the sum 7. Hence, the prob of that and A is always 1/36. And the prob of that _one_ event is of course 1/6. Hence, they are all independent

35.
Let's say the first person has a birthday on day 1. For the other person to have the same birthday, he has only one choice - day 1. 1 out of 365 possible days - 1/365

37.
(a) P(current flowing) = P(both relays closing) = P(A). P(B) = 0.8 * 0.8 = 0.64

(b) P(current flowing) = P(first relay closing) or P(second relay closing) = P(A) + P(B) + P(A∩B) = 0.8 + 0.8 - 0.64 = 0.96

(c) P(current flowing) = P(both upper relays close) or P(both lower relays close)
= (P(upper1).P(upper2)) or (P(upper1).P(upper2))
= 0.8 * 0.8 + 0.8 * 0.8 - (0.8 * 0.8 * 0.8 * 0.8) = 0.8704

38.
No, they are not independent.
Intuitively, P(B) changes once A has happened.


39.
Yes, they will be. In this case, P(A) is alway 1/2, and so is P(B)
P(A∩B) = 1/4 = P(A).P(B)

40.
(a) P(all 4 yes or all 4 no) = 0.7 * 0.7 * 0.7 * 0.7 + 0.3 * 0.3 * 0.3 * 0.3 = 0.2401 + 0.0081 = 0.2482

(b) P(first 2 no, last 2 yes) = 0.9 * 0.49 = 0.441

(c) P(atleast one no) = 1 - P(no nos) = 1 - (all yes) = 1 - 0.2482 = 0.7518

(d) 4 cases where exactly three answer yes.
P(3 yes) = (0.7 cubed * 0.3) quadrupled = 0.000112114

41.
(a) Rain on all 3 = 4/31 * 3/30 * 16/31 = 0.006659729
(b) Dry on all 3 = (1-4/31) * (1-3/30) * (1-16/31) = 0.379292404
(c) P(Ph, Mo, not LA) = 4/31 * (1-3/30) * 16/31 = 0.059937565
(d) ... same stuff.
(f) Sum of c,d,e

42.
(a) P(both are defective) = P(A).P(B) = 0.005
(b) P(both are not defective) = (1-P(A)). P(1-P(B)) = 0.0855
(c) P(exactly one) = P(A).P(not B) + P(not A).P(B) = 0.1 * 0.95 + 0.9 * 0.05 = 0.14
(d) P(A|defective) = P(A and defective) / P(defective)
= 0.10/(0.10 + 0.05) = 0.666666667
(e) P(B|defective) = P(B and defective) / P(defective)
= 0.05/(0.10 + 0.05) = 0.333333333

43.
(a) P(CF) = 1/2 + 1/2 = 1/4 (one from each parent)
(b) Q Not clear.