Conditional probability measures the likelihood that an event \(A\) happens given that another event B has already happened, denoted by \(\mathrm{P}\left( A|B \right)\). It focuses only on outcomes where \(B\) is true, then checks how often \(A\) also occurs within those cases. For \(P(B)>0\), \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\). For independent events, \(P(A\mid B)=P(A)\) when \(P(B)>0\), and \(P(B\mid A)=P(B)\) when \(P(A)>0\). Total Probability helps find the probability of an event \(A\) by breaking it into several mutually exclusive scenarios \(A\cap {{B}_{1}}, A\cap {{B}_{2}}, \ldots A\cap {{B}_{n}}\), where \({{B}_{1}}, {{B}_{2}}, \ldots {{B}_{n}}\) are mutually exclusive to one another and exhaustive. It combines the contributions to event \(A\) from the disjoint, exhaustive cases \(B_i\). • Formula: \(P(A)=\sum\limits_{i=1}^{n}{\mathrm{P}\left( A\cap {{B}_{i}} \right)}=\sum\limits_{i=1}^{n}{\mathrm{P}\left( A|{{B}_{i}} \right)\mathrm{P}\left( {{B}_{i}} \right)}\) • A simpler case: \(P(A)=\mathrm{P}\left( A\cap B \right)+\mathrm{P}\left( A\cap B' \right)\) Bayes’ Rule helps us reverse conditional probabilities. It answers questions like: “Given that event \(A\) has occurred, what is the probability that event \(B\) was the cause?” It updates probability when new information is revealed. For \(P(A)>0\) and \(P(B)>0\), \(P(B\mid A)=\frac{P(A\mid B)P(B)}{P(A)}\). Probability Diagram Probability of \(A\), \(\mathrm{P}\left( A \right)=\) Probability of \(A'\), \(\mathrm{P}\left( A' \right)=\) Probability of \(A\cap B\), \(\mathrm{P}\left( A\cap B \right)=\) Probability of \(A'\cap B'\), \(\mathrm{P}\left( A'\cap B' \right)=\) Probability of \(B\) given that \(A\) happens, \(\mathrm{P}\left( B|A \right)=\) Probability of \(B'\) given that \(A\) happens, \(\mathrm{P}\left( B'|A \right)=\) Probability of \(B\) given that \(A'\) happens, \(\mathrm{P}\left( B|A' \right)=\) Probability of \(B'\) given that \(A'\) happens, \(\mathrm{P}\left( B'|A' \right)=\) Probability of \(B\), \(\mathrm{P}\left( B \right)=\) Probability of \(B'\), \(\mathrm{P}\left( B' \right)=\) \(A\) \(A'\) \(B\) \(w\) \(x\) \(B'\) \(y\) \(z\). In conditional expressions, the conditioning event must have positive probability. For the partition formula, take \(P(B_i)>0\). The table entries \(w,x,y,z\) are nonnegative joint probabilities with sum 1; each displayed conditional ratio requires a positive denominator.

Conditional Probability, Total Probability and Bayes’ Rule

IP 4

Conditional probability measures the likelihood that an event \(A\) happens given that another event B has already happened, denoted by \(\mathrm{P}\left( A|B \right)\). It focuses only on outcomes where \(B\) is true, then checks how often \(A\) also occurs within those cases.

For \(P(B)>0\), \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\).

For independent events, \(P(A\mid B)=P(A)\) when \(P(B)>0\), and \(P(B\mid A)=P(B)\) when \(P(A)>0\).

Total Probability helps find the probability of an event \(A\) by breaking it into several mutually exclusive scenarios \(A\cap {{B}_{1}}, A\cap {{B}_{2}}, \ldots A\cap {{B}_{n}}\), where \({{B}_{1}}, {{B}_{2}}, \ldots {{B}_{n}}\) are mutually exclusive to one another and exhaustive. It combines the contributions to event \(A\) from the disjoint, exhaustive cases \(B_i\).

• Formula: \(P(A)=\sum\limits_{i=1}^{n}{\mathrm{P}\left( A\cap {{B}_{i}} \right)}=\sum\limits_{i=1}^{n}{\mathrm{P}\left( A|{{B}_{i}} \right)\mathrm{P}\left( {{B}_{i}} \right)}\)

• A simpler case: \(P(A)=\mathrm{P}\left( A\cap B \right)+\mathrm{P}\left( A\cap B' \right)\)

Bayes’ Rule helps us reverse conditional probabilities. It answers questions like: “Given that event \(A\) has occurred, what is the probability that event \(B\) was the cause?”
It updates probability when new information is revealed.

For \(P(A)>0\) and \(P(B)>0\), \(P(B\mid A)=\frac{P(A\mid B)P(B)}{P(A)}\).

Probability Diagram

Probability of \(A\), \(\mathrm{P}\left( A \right)=\)
Probability of \(A'\), \(\mathrm{P}\left( A' \right)=\)

Probability of \(A\cap B\), \(\mathrm{P}\left( A\cap B \right)=\)
Probability of \(A'\cap B'\), \(\mathrm{P}\left( A'\cap B' \right)=\)

Probability of \(B\) given that \(A\) happens, \(\mathrm{P}\left( B|A \right)=\)
Probability of \(B'\) given that \(A\) happens, \(\mathrm{P}\left( B'|A \right)=\)
Probability of \(B\) given that \(A'\) happens, \(\mathrm{P}\left( B|A' \right)=\)
Probability of \(B'\) given that \(A'\) happens, \(\mathrm{P}\left( B'|A' \right)=\)

Probability of \(B\), \(\mathrm{P}\left( B \right)=\)
Probability of \(B'\), \(\mathrm{P}\left( B' \right)=\)

	\(A\)	\(A'\)
\(B\)	\(w\)	\(x\)
\(B'\)	\(y\)	\(z\).

In conditional expressions, the conditioning event must have positive probability. For the partition formula, take \(P(B_i)>0\). The table entries \(w,x,y,z\) are nonnegative joint probabilities with sum 1; each displayed conditional ratio requires a positive denominator.

Example

Machine \(X\) makes \(60\%\) of a factory's bolts and machine \(Y\) makes the rest. \(3\%\) of \(X\)'s bolts and \(5\%\) of \(Y\)'s bolts are defective. A bolt is chosen at random and is defective (\(D\)).

\(P(D)=0.6(0.03)+0.4(0.05)=0.018+0.020=0.038\)

\(P(Y\mid D)=\dfrac{P(Y\cap D)}{P(D)}=\dfrac{0.020}{0.038}=\dfrac{10}{19}\approx0.526\)

Exam check: the denominator in Bayes' rule is the total probability of the observed event, found by adding every path that leads to it.

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