Lesson 5: Independent Events
Lesson 5: Independent EventsOverview
In this lesson, we learn what it means for two (or more) events to be independent. We'll formally learn, for example, why we say that the outcome of the flip of a fair coin is independent of the flips that came before it.
Objectives
 Learn how to determine if two events are independent.
 Learn how to find the probability of the intersection of two events if the two events are independent.
 Learn how to determine if three or more events are pairwise independent.
 Learn how to determine if three or more events are mutually independent.
 Understand each step of the proofs contained in the lesson.
 Practice applying the techniques learned in the lesson to new problems.
5.1  Two Definitions
5.1  Two DefinitionsExample 51
A couple plans to have three children. What is the probability that the second child is a girl? And, what is the probability that the second child is a girl given that the first child is a girl?
This example leads us to a formal definition of independent events.
 Independent Events

Events \(A\) and \(B\) are independent events if the occurrence of one of them does not affect the probability of the occurrence of the other. That is, two events are independent if either:
\(P(BA)=P(B)\)
(provided that \(P(A)>0\)) or:
\(P(AB=P(A))\)
(provided that \(P(B)>0\)).
Now, since independence tells us that \(P(BA)=P(B)\), we can substitute \(P(B)\) in for \(P(BA)\) in the formula given to us by the multiplication rule:
\(P(A\cap B)=P(A)\times P(BA)\)
yielding:
\(P(A\cap B)=P(A)\times P(B)\)
This substitution leads us to an alternative definition of independence.
 Independent Events

Events \(A\) and \(B\) are independent events if and only if :
\(P(A\cap B)=P(A)\times P(B)\)
Otherwise, \(A\) and \(B\) are called dependent events.
Recall that the "if and only if" (often written as "iff") in that definition means that the ifthen statement works in both directions. That is, the definition tells us two things:
 If events \(A\) and \(B\) are independent, then \(P(A\cap B)=P(A)\times P(B)\).
 If \(P(A\cap B)=P(A)\times P(B)\), then events \(A\) and \(B\) are independent.
The next example illustrates the first of these two directions, while the second example illustrates the second direction.
Example 52
A recent survey of students suggested that 10% of Penn State students commute by bike, while 40% of them have a significant other. Based on this survey, what percentage of Penn State students commute by bike and have a significant other?
Answer
Let's let \(B\) be the event that a randomly selected Penn State student commutes by bike and \(S\) be the event that a randomly selected Penn State student has a significant other. If \(B\) and \(S\) are independent events (okay??), then the definition tells us that:
\(P(B\cap S)=P(B)\times P(S)=0.10(0.40)=0.04\)
That is, 4% of Penn State students commute by bike and have a significant other. (Is this result at all meaningful??)
Example 53
Let's return to the couple that plans to have three children. Is the event that the couple's third child is a girl independent of the event that the couple's first two children are girls?
Answer
Again, letting \(G\) denote girl and \(B\) denote boy, the sample space of the genders of the couple's three children looks like this:
\(\{GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB\}\)
Let \(C\) be the event that the couple's first two children are girls, and let \(D\) be the event that the third child is a girl. Then event \(C\) looks like this:
\(\{GGG, GGB\}\)
where the first two children are restricted to be girls (\(G\)), while the third child could be either a girl (\(G\)) or a boy (\(B\)). For event \(D\), there are no restrictions on the first two children, but the third child must be a girl. Therefore, event \(D\) looks like this:
\(\{GGG, BBG, BGG, GBG\}\)
Now, \(C\cap D\) is the event that all three children are girls and hence it looks like:
\(\{GGG\}\)
Using the classical approach to assigning a probability to the three events \(C, D\) and \(C\cap D\):
 \(P(C) = \dfrac{2}{8}\)
 \(P(D) = \dfrac{4}{8}\)
 \(P(C\cap D) = \dfrac{1}{8}\)
Now, since:
\(P(C)\times P(D)=\dfrac{2}{8}\left(\dfrac{4}{8}\right)=\dfrac{8}{64}=\dfrac{1}{8}=P(C\cap D)\)
we can conclude that events \(C\) and \(D\) are independent. That is, the event that the couple's third child is a girl is independent of the event that the first two children were girls. This result seems quite intuitive, eh? If so, then it is quite interesting that so many people fall prey to the "Gambler's Fallacy" illustrated in the next example.
Example 54
A man tosses a fair coin eight times and observes whether the toss yields a head (H) or a tail (T) on each toss. Which of the following sequences of coin tosses is the man more likely to get a head (H) on his next toss? This one:
\( TTTTTTTT\)
or this one:
\(HHTHTTHH\)
The answer is neither as illustrated here:
The moral of the story is to be careful not to fall prey to "Gambler's Fallacy," which occurs when a person mistakenly assumes that a departure from what occurs on average in the long term will be corrected in the short term. In this case, a person would be mistaken to think that just because the coin was departing from the average (half the tosses being heads and half the tosses being tails) by getting eight straight tails in row that a head was due to be tossed. A classic example of Gambler's Fallacy occurred on August 18, 1913 at the casino in Monte Carlo, in which:
A joke told among statisticians demonstrates the nature of the fallacy. A man attempts to board an airplane with a bomb. When questioned by security, he explains that he was just trying to protect himself: "The chances of an airplane having a bomb on it are very small," he reasons, "and certainly the chances of having two are almost none!" A similar example is in the book by John Irving called The World According to Garp, in which the hero Garp decides to buy a house immediately after a small plane crashes into it, reasoning that the chances of another plane hitting the house have just dropped to zero.
5.2  Three Theorems
5.2  Three TheoremsOn this page, we present, prove, and then use three sometimes helpful theorems.
Example 55
A nationwide poll determines that 72% of the American population loves eating pizza. If two people are randomly selected from the population, what is the probability that the first person loves eating pizza, while the second one does not?
Answer
Let \(A\) be the event that the first person loves pizza, and let \(B\) be the event that the second person loves pizza. Because the two people are selected randomly from the population, it is reasonable to assume that \(A\) and \(B\) are independent events. If \(A\) and \(B\) are independent events, then \(A\) and \(B^\prime\) are also independent events. Therefore, \(P(A\cap B^\prime)\), the probability that the first person loves eating pizza, while the second one does not can be calculated by multiplying their individual probabilities together. That is, the probability that the first person loves eating pizza, while the second one does not is:
\(P(A\cap B^\prime)=0.72(10.72)=0.202\)
5.3  Mutual Independence
5.3  Mutual IndependenceExample 56
Consider a roulette wheel that has 36 numbers colored red (\(R\)) or black (\(B\)) according to the following pattern:
1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18 
R  R  R  R  R  B  B  B  B  R  R  R  R  B  B  B  B  B 
36  35  34  33  32  31  30  29  28  27  26  25  24  23  22  21  20  19 
and define the following three events:
 Let \(A\) be the event that a spin of the wheel yields a RED number = \(\{1, 2, 3, 4, 5, 10, 11, 12, 13, 24, 25, 26, 27, 32, 33, 34, 35, 36\}\).
 Let \(B\) be the event that a spin of the wheel yields an EVEN number = \(\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36\}\).
 Let \(C\) be the event that a spin of the wheel yields a number no greater than 18 = \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\}\).
Now consider the following two questions:
 Are the events \(A\), \(B\), and \(C\) "pairwise independent?" That is, is event \(A\) independent of event \(B\); event \(A\) independent of event \(C\); and \(B\) independent of event \(C\)?
 Does \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\)?
Let's take a look:
So... this example illustrates that something seems to be lacking for the complete independence of events \(A, B, \text{ and }C\). And, that's why the second condition exists in the following definition.
 Mutually Independent Event

Three events \(A, B,\text{ and }C\) are mutually independent if and only if the following two conditions hold:
 The events are pairwise independent. That is,
 \(P(A\cap B)=P(A)\times P(B)\) and...
 \(P(A\cap C)=P(A)\times P(C)\) and...
 \(P(B\cap C)=P(B)\times P(C)\)
 \(P(A\cap B\cap C)=P(A)\times P(B)\times P(C)\)
 The events are pairwise independent. That is,
The idea of mutual independence can be extended to four or more events — each pair, triple, quartet, and so on, must satisfy the above type of multiplication rule.
Example 57
One ball is drawn randomly from a bowl containing four balls numbered 1, 2, 3, and 4. Define the following three events:
 Let \(A\) be the event that a 1 or 2 is drawn. That is, \(A=\{1, 2\}\).
 Let \(B\) be the event that a 1 or 3 is drawn. That is, \(B = \{1, 3\}\).
 Let \(C\) be the event that a 1 or 4 is drawn. That is, \(C = \{1, 4\}\).
Are events \(A, B,\text{ and }C\) pairwise independent? Are they mutually independent?
This example illustrates, as does the previous example, that pairwise independence among three events does not necessarily imply that the three events are mutually independent.
Example 58
A pair of fair sixsided dice is tossed once yielding the following sample space:
\(S=\left\{\begin{array}{l}{(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)} \\ {(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)} \\ {(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)} \\ {(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)} \\ {(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)} \\ {(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}\end{array}\right.\)
Define the following three events:
 Let \(A\) be the event that the first die is a 1, 2, or 3.
 Let \(B\) be the event that the first die is a 3, 4, or 5.
 Let \(C\) be the event that the sum of the two faces equals 9. That is \(C = \{(3,6), (6,3), (4,5), (5,4)\}\).
Are events \(A, B,\text{ and }C\) pairwise independent? Are they mutually independent?
In solving that problem, I admit to being a little looseygoosey with the definition of "mutual independence." That's why I said "a sort of mutual independence." Now that I haven't been perfectly clear so far, let me set the record straight by being perfectly clear. This example illustrates that the second condition of mutual independence among the three events \(A, B,\text{ and }C\) (that is, the probability of the intersection of the three events equals the probabilities of the individual events multiplied together) does not necessarily imply that the first condition of mutual independence holds (that is, three events \(A, B,\text{ and }C\) are pairwise independent). In order to check for mutual independence, you clearly need to check both the first and second conditions.
5.4  A Closing Example
5.4  A Closing ExampleExample 59
A zealous father is trying to advance his son Jake's promising tennis career. Jake's father tells Jake that he'll give him \$500 if he wins (at least) two tennis sets in a row in a threeset series to be played with his father and the club champion alternately. The champion is a better player than Jake's father. Jake can choose who he plays first. If he plays his father first, he plays his father twice... first father, then champion, then father. If he plays the champion first, he plays his father once... first champion, then father, then champion. Which threeset series should Jake choose so as to maximize his chances of winning the \$500?
Solution
Because the champion plays better than the father, it seems reasonable that fewer sets should be played with the champion. On the other hand, the middle set is the key one, because Jake cannot get two wins in a row without winning the middle set. Now, let's define the following:
 Let \(C\)stand for champion.
 Let \(F\) stand for father.
 Let \(W\) denote the event that Jake wins a set. Let \(f\) denote the probability that Jake wins any set from his father. Let \(c\) denote the probability that Jake wins any set from the champion.
 Let \(L\) denote the event that Jake loses a set.
Let's first consider the case where Jake plays in \(FCF\) order (his father first, then the champion, then his father):
Now, let's consider the case where Jake plays in \(CFC\) order (the champion first, then his father, then the champion):
In summary, the probability that Jake wins at least two sets in a row if he plays in \(FCF\) order, is \(fc(2f)\), and the probability that Jake wins at least two sets in a row if he plays in \(CFC\) order is \(fc(2c)\). For example, if \(f=0.8\) and \(c=0.4\), then the probability that Jake wins if he plays in \(FCF\) order is 0.384, while the probability that Jake wins if he plays in \(CFC\) order is 0.512. Now, because Jake is more likely to beat his father than to beat the champion, \(f\) is larger than \(c\), and \(2f\) is smaller than \(2c\). Therefore, in general, \(fc(2c)\) is larger than \(fc(2f)\). That is, Jake is more likely to win the \$500 if he plays the champion first, his father second, and the champion again. As such, it appears that the importance of winning the middle game outweighs the disadvantage of playing the champion twice!