diff --git a/source/sec-Independent-Events.ptx b/source/sec-Independent-Events.ptx
index 6ca3cee..7b8028b 100644
--- a/source/sec-Independent-Events.ptx
+++ b/source/sec-Independent-Events.ptx
@@ -2,21 +2,26 @@
Independent Events
- The idea of conditional probability is that knowledge of one event can change our understanding of the probability of another.
+ The idea of conditional probability is that knowledge of one event can
+ change our understanding of the probability of another.
But it's also important to understand when this is not the case.
- Events A, B are independent if \Pr(A \mid B) = \Pr(A).
+ Events A, B are independent if
+ \Pr(A \mid B) = \Pr(A).
- The equation \Pr(A\mid B) = \Pr(A) very concretely says: knowledge that the event B has occurred does not change our understanding of \Pr(A).
- Assuming \Pr(A) \neq 0, this is equivalent to \Pr(A\cap B) = \Pr(A)\Pr(B), since:
+ The equation \Pr(A\mid B) = \Pr(A) very concretely says: knowledge
+ that the event B has occurred does not change our understanding of
+ \Pr(A).
+ Assuming \Pr(A) \neq 0, this is equivalent to
+ \Pr(A\cap B) = \Pr(A)\Pr(B), since:
\Pr(A\mid B) \amp = \Pr(A)
\frac{\Pr(A\cap B)}{\Pr(B)} \amp = \Pr(A)
@@ -28,12 +33,17 @@
- In , we considered an experiment in which we rolled a fair 6-sided die twice.
- We defined events A that the sum of the rolls is at least 10 and B that the first roll is a 6, and we found that \Pr(A \mid B) \neq \Pr(A), so A and B are not independent.
+ In , we considered an experiment
+ in which we rolled a fair 6-sided die twice.
+ We defined events A that the sum of the rolls is at least 10 and
+ B that the first roll is a 6, and we found that
+ \Pr(A \mid B) \neq \Pr(A), so A and B are not
+ independent.
- Now consider the event C that the sum of the rolls is 7, which sounds very similar to the event A.
+ Now consider the event C that the sum of the rolls is 7, which
+ sounds very similar to the event A.
We have:
C \amp = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}
@@ -42,129 +52,192 @@
\Pr(B\cap C) \amp = \frac{1}{36}
\text{therefore: } \Pr(C\mid B) \amp = \frac{\Pr(C\cap B)}{\Pr(B)} = \frac{1/36}{1/6} = \frac{1}{6} = \Pr(C)
- So, even though the descriptions of events A and C are very similar, the event C is independent with B, while A is not.
+ So, even though the descriptions of events A and C are
+ very similar, the event C is independent with B, while
+ A is not.
-
-
-
-
- An experiment consists of rolling a fair die two times.
- Let A be the event that the sum is even, and let B be the event that the second roll is higher than the first.
- Are A and B independent?
-
-
+
+
+
+ An experiment consists of rolling a fair die two times.
+ Let A be the event that the sum is even, and let B be
+ the event that the second roll is higher than the first.
+ Are A and B independent?
+
+
-
-
-
- A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6),
- \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6),
- \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\}
- B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
- \amp (2, 3), (2, 4), (2, 5), (2, 6),
- \amp (3, 4), (3, 5), (3, 6),
- \amp (4, 5), (4, 6),
- \amp (5, 6)\}
- A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\}
-
- So \Pr(A) = \frac{18}{36} = \frac{1}{2}, \Pr(B) = \frac{15}{36} = \frac{5}{12}, and \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}.
- Finally:
-
- \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A),
-
- so A and B are not independent.
-
-
-
+
+
+ A and B are not independent.
+
+
-
-
-
- An experiment consists of flipping a fair coin three times.
- Let A be the event that the first and second flips match.
- Let B be the event that there are at least two heads.
- Are A and B independent?
-
-
+
+
+
+ A = \{ \amp (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6),
+ \amp (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6),
+ \amp (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)\}
+ B = \{ \amp (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
+ \amp (2, 3), (2, 4), (2, 5), (2, 6),
+ \amp (3, 4), (3, 5), (3, 6),
+ \amp (4, 5), (4, 6),
+ \amp (5, 6)\}
+ A \cap B = \{ \amp (1, 3), (1, 5), (2, 4), (2, 6), (3, 5), (4, 6)\}
+
+ So \Pr(A) = \frac{18}{36} = \frac{1}{2},
+ \Pr(B) = \frac{15}{36} = \frac{5}{12}, and
+ \Pr(A\cap B) = \frac{6}{36} = \frac{1}{6}.
+ Finally:
+
+ \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{6/36}{15/36} = \frac{6}{15} = \frac{2}{5} \neq \Pr(A),
+
+ so A and B are not independent.
+
+
+
-
-
-
- A \amp = \{ HHH, HHT, TTH, TTT \}
- B \amp = \{ HHH, HHT, HTH, THH \}
- A\cap B \amp = \{HHH, HHT\}
-
- So \Pr(A) = \frac{4}{8} = \frac{1}{2}, \Pr(B) = \frac{4}{8} = \frac{1}{2}, and \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}.
- Finally:
-
- \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A),
-
- so A and B are independent.
-
-
-
+
+
+
+ An experiment consists of flipping a fair coin three times.
+ Let A be the event that the first and second flips match.
+ Let B be the event that there are at least two heads.
+ Are A and B independent?
+
+
-
-
-
- Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the sample space \Omega = \{1, 2, 3, 4, 5, 6\}.
- Create a probability distribution for \Omega so that A, B are independent.
-
-
+
+
+ A and B are independent.
+
+
-
-
- Example Distribution
+
+
+
+ A \amp = \{ HHH, HHT, TTH, TTT \}
+ B \amp = \{ HHH, HHT, HTH, THH \}
+ A\cap B \amp = \{HHH, HHT\}
+
+ So \Pr(A) = \frac{4}{8} = \frac{1}{2},
+ \Pr(B) = \frac{4}{8} = \frac{1}{2}, and
+ \Pr(A\cap B) = \frac{2}{8} = \frac{1}{4}.
+ Finally:
+
+ \Pr(A \mid B) \amp = \frac{\Pr(A\cap B)}{\Pr(B)} = \frac{2/8}{4/8} = \frac{2}{4} = \frac{1}{2} = \Pr(A),
+
+ so A and B are independent.
+
+
+
-
-
- | x |
- \Pr(x) |
-
+
+
+
+ Let A = \{1, 2, 3\} and B = \{3, 4, 5\} be events in the
+ sample space \Omega = \{1, 2, 3, 4, 5, 6\}.
+ Create a probability distribution for \Omega so that
+ A, B are independent.
+
+
-
- | 1 |
- 0.1 |
-
+
+
+ Example Distribution
-
- | 2 |
- 0.2 |
-
+
+
+ | x |
+ \Pr(x) |
+
-
- | 3 |
- 0.2 |
-
+
+ | 1 |
+ 0.1 |
+
-
- | 4 |
- 0.1 |
-
+
+ | 2 |
+ 0.2 |
+
-
- | 5 |
- 0.1 |
-
+
+ | 3 |
+ 0.2 |
+
-
- | 6 |
- 0.3 |
-
-
-
+
+ | 4 |
+ 0.1 |
+
-
- Now \Pr(A) = 0.5, \Pr(B) = 0.4, and
-
- \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B),
-
- so A and B are independent.
-
-
-
+
+ | 5 |
+ 0.1 |
+
+
+
+ | 6 |
+ 0.3 |
+
+
+
+
+
+
+
+ Example Distribution
+
+
+
+ | x |
+ \Pr(x) |
+
+
+
+ | 1 |
+ 0.1 |
+
+
+
+ | 2 |
+ 0.2 |
+
+
+
+ | 3 |
+ 0.2 |
+
+
+
+ | 4 |
+ 0.1 |
+
+
+
+ | 5 |
+ 0.1 |
+
+
+
+ | 6 |
+ 0.3 |
+
+
+
+
+
+ Now \Pr(A) = 0.5, \Pr(B) = 0.4, and
+
+ \Pr(A\cap B) = 0.2 = (0.5)(0.4) = \Pr(A)\Pr(B),
+
+ so A and B are independent.
+
+
+
\ No newline at end of file