# 5.1 - Two Definitions

5.1 - Two Definitions

## Example 5-1

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:

$$P(B|A)=P(B)$$

(provided that $$P(A)>0$$) or:

$$P(A|B=P(A))$$

(provided that $$P(B)>0$$).

Now, since independence tells us that $$P(B|A)=P(B)$$, we can substitute $$P(B)$$ in for $$P(B|A)$$ in the formula given to us by the multiplication rule:

$$P(A\cap B)=P(A)\times P(B|A)$$

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 if-then statement works in both directions. That is, the definition tells us two things:

1. If events $$A$$ and $$B$$ are independent, then $$P(A\cap B)=P(A)\times P(B)$$.
2. 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 5-2

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?

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 5-3

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?

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 5-4

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:

... black came up a record twenty-six times in succession [in roulette]. … [There] was a near-panicky rush to bet on red, beginning about the time black had come up a phenomenal fifteen times. In application of the maturity [of the chances] doctrine, players doubled and tripled their stakes, this doctrine leading them to believe after black came up the twentieth time that there was not a chance in a million of another repeat. In the end the unusual run enriched the Casino by some millions of francs.   [Source: Darrell Huff & Irving Geis, How to Take a Chance (1959), pp. 28-29.]

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.

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