3.2: Conditional Probability

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Page ID: 47233

Franz S. Hover & Michael S. Triantafyllou
Massachusetts Institute of Technology via MIT OpenCourseWare

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If a composite event \(M\) is known to have occurred, a question arises as to the probability that one of the constituent simple events Ai occurred. This is written as \(P(A_j |M)\), read as ”the probability of \(A_j\), given \(M\),” and this is a conditional probability. The key concept here is that \(M\) replaces \(S\) as the event space, so that \(p(M) = 1\). This will have the natural effect of inflating the probabilities of events that are part of event \(M\), and in fact

\[p(A_j | M) = \dfrac{p(A_j \cap M)}{p(M)}. \]

Referring to our die example above, if \(M\) is the event of an even result, then we have

\begin{align*} M = A_2 \cup A_4 \cup A_6 \\[4pt] p(M \cap A_2) = p(A_2) = 1/6 \\[4pt] p(M) = 1/2 \longrightarrow \\[4pt] p(A_2 | M) = \dfrac{1/6}{1/2} = 1/3. \end{align*}

Given that an event result was observed (composite event \(M\)), the probability that a two was rolled is 1/3. Now if all the \(A_j\) are independent (simple) events and \(M\) is a composite event, then we can write an opposing rule:

\[ p(M) = p(M|A_1) \, p(A_1) \, + \, . \, . \, . + \, p(M|A_n) \, p(A_n). \]

This relation collects conditional probabilities of \(M\) given each separate event \(A_i\). Its logic is easily seen in a graph. Here is an example of how to use it in a practical problem. Box A has 2000 items in it of which 5% are defective; box B has 500 items with 40% defective; boxes C and D each contain 1000 items with 10% defective. If a box is picked at random, and one item is taken from that box, what is the probability that it is defective? \(M\) is the composite event of a defective item, so we are after \(p(M)\). We apply the formula above to find

\[ p(M) = 0.05 \times 0.25 + 0.40 \times 0.25 + 0.10 \times 0.25 + 0.10 \times 0.25 = 0.1625. \nonumber \]