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4.2: Conditional Probability

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    We have described the whole foundation of the theory of probability as coming from imperfect knowledge, in the sense that we don’t know for sure if an event \(A\) will happen any particular time we do the experiment but we do know, in the long run, in what fraction of times \(A\) will happen. Or, at least, we claim that there is some number \(P(A)\) such that after running the experiment \(N\) times, out of which \(n_A\) of these times are when \(A\) happened, \(P(A)\) is approximately \(n_A/N\) (and this ratio gets closer and closer to \(P(A)\) as \(N\) gets bigger and bigger).

    But what if we have some knowledge? In particular, what happens if we know for sure that the event \(B\) has happened – will that influence our knowledge of whether \(A\) happens or not? As before, when there is randomness involved, we cannot tell for sure if \(A\) will happen, but we hope that, given the knowledge that \(B\) happened, we can make a more accurate guess about the probability of \(A\).

    [eg:condprob1] If you pick a person at random in a certain country on a particular date, you might be able to estimate the probability that the person had a certain height if you knew enough about the range of heights of the whole population of that country. [In fact, below we will make estimates of this kind.] That is, if we define the event \[A=\text{``the random person is taller than 1.829 meters (6 feet)''}\] then we might estimate \(P(A)\).

    But consider the event \[B=\text{``the random person's parents were both taller than 1.829 meters''}\ .\] Because there is a genetic component to height, if you know that \(B\) happened, it would change your idea of how likely, given that knowledge, that \(A\) happened. Because genetics are not the only thing which determines a person’s height, you would not be certain that \(A\) happened, even given the knowledge of \(B\).

    Let us use the frequentist approach to derive a formula for this kind of probability of \(A\) given that \(B\) is known to have happened. So think about doing the repeatable experiment many times, say \(N\) times. Out of all those times, some times \(B\) happens, say it happens \(n_B\) times. Out of those times, the ones where \(B\) happened, sometimes \(A\) also happened. These are the cases where both \(A\) and \(B\) happened – or, converting this to a more mathematical descriptions, the times that \(A\cap B\) happened – so we will write it \(n_{A\cap B}\).

    We know that the probability of \(A\) happening in the cases where we know for sure that \(B\) happened is approximately \(n_{A\cap B}/n_B\). Let’s do that favorite trick of multiplying and dividing by the same number, so finding that the probability in which we are interested is approximately \[\frac{n_{A\cap B}}{n_B} = \frac{n_{A\cap B}\cdot N}{N\cdot n_B} = \frac{n_{A\cap B}}{N}\cdot\frac{N}{n_B} = \frac{n_{A\cap B}}{N} \Bigg/ \frac{n_B}{N} \approx P(A\cap B) \Big/ P(B)\]

    Which is why we make the

    [def:condprob] The conditional probability is \[P(A|B) = \frac{P(A\cap B)}{P(B)}\ .\] Here \(P(A|B)\) is pronounced the probability of \(A\) given \(B\).

    Let’s do a simple

    EXAMPLE 4.2.3. Building off of Example 4.1.19, note that the probability of rolling a \(2\) is \(P(\{2\})=1/6\) (as is the probability of rolling any other face – it’s a fair die). But suppose that you were told that the roll was even, which is the event \(\{2, 4, 6\}\), and asked for the probability that the roll was a \(2\) given this prior knowledge. The answer would be \[P(\{2\}\mid\{2, 4, 6\})=\frac{P(\{2\}\cap\{2, 4, 6\})}{P(\{2, 4, 6\})} =\frac{P(\{2\})}{P(\{2, 4, 6\})} = \frac{1/6}{1/2} = 1/3\ .\] In other words, the probability of rolling a \(2\) on a fair die with no other information is \(1/6\), which the probability of rolling a \(2\) given that we rolled an even number is \(1/3\). So the probability doubled with the given information.

    Sometimes the probability changes even more than merely doubling: the probability that we rolled a \(1\) with no other knowledge is \(1/6\), while the probability that we rolled a \(1\) given that we rolled an even number is \[P(\{1\}\mid\{2, 4, 6\})=\frac{P(\{1\}\cap\{2, 4, 6\})}{P(\{2, 4, 6\})} =\frac{P(\emptyset)}{P(\{2, 4, 6\})} = \frac{0}{1/2} = 0\ .\]

    But, actually, sometimes the conditional probability for some event is the same as the unconditioned probability. In other words, sometimes knowing that \(B\) happened doesn’t change our estimate of the probability of \(A\) at all, they are no really related events, at least from the point of view of probability. This motivates the

    [def:independent] Two events \(A\) and \(B\) are called independent if \(P(A\mid B)=P(A)\).

    Plugging the defining formula for \(P(A\mid B)\) into the definition of independent, it is easy to see that

    FACT 4.2.5. Events \(A\) and \(B\) are independent if and only if \(P(A\cap B)=P(A)\cdot P(B)\).

    EXAMPLE 4.2.6. Still using the situation of Example 4.1.19, we saw in Example 4.2.3 that the events \(\{2\}\) and \(\{2, 3, 4\}\) are not independent since \[P(\{2\}) = 1/6 \neq 1/3 = P(\{2\}\mid\{2, 4, 6\})\] nor are \(\{1\}\) and \(\{2, 3, 4\}\), since \[P(\{1\}) = 1/6 \neq 0 = P(\{1\}\mid\{2, 4, 6\})\ .\] However, look at the events \(\{1, 2\}\) and \(\{2, 4, 6\}\): \[\begin{aligned} P(\{1, 2\}) = P(\{1\}) + P(\{2\}) &= 1/6 + 1/6\\ &= 1/3\\ &= \frac{1/6}{1/2}\\ &= \frac{P(\{1\})}{P(\{2, 4, 6\})}\\ &= \frac{P(\{1, 2\}\cap\{2, 4, 6\})}{P(\{2, 4, 6\})}\\ &= P(\{1, 2\}\mid\{2, 4, 6\})\end{aligned}\] which means that they are independent!

    EXAMPLE 4.2.7. We can now fully explain what was going on in Example 4.1.21. The two fair dice were supposed to be rolled in a way that the first roll had no effect on the second – this exactly means that the dice were rolled independently. As we saw, this then means that each individual outcome of sample space \(S\) had probability \(\frac{1}{36}\). But the first roll having any particular value is independent of the second roll having another, e.g., if \(A=\{11, 12, 13, 14, 15, 16\}\) is the event in that sample space of getting a \(1\) on the first roll and \(B=\{14, 24, 34, 44, 54, 64\}\) is the event of getting a \(4\) on the second roll, then events \(A\) and \(B\) are independent, as we check by using Fact 4.2.5: \[\begin{aligned} P(A\cap B) &= P(\{14\})\\ &= \frac{1}{36}\\ &= \frac16\cdot\frac16\\ &= \frac{6}{36}\cdot\frac{6}{36}\\ &=P(A)\cdot P(B)\ .\end{aligned}\] On the other hand, the event “the sum of the rolls is \(4\),” which is \(C=\{13, 22, 31\}\) as a set, is not independent of the value of the first roll, since \(P(A\cap C)=P(\{13\})=\frac{1}{36}\) but \(P(A)\cdot P(C)=\frac{6}{36}\cdot\frac{3}{36}=\frac16\cdot\frac{1}{12}=\frac{1}{72}\).

    This page titled 4.2: Conditional Probability is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jonathan A. Poritz via source content that was edited to the style and standards of the LibreTexts platform.