10.3 Mutually exclusive events
Definition 10.10. Two events \(A\) and \(B\) from the same sample space are mutually exclusive if they have no outcomes in common, so that they cannot both happen. Then \[P(A\cap B)=0.\]
Drawing a red card and drawing a club are mutually exclusive, because every club is black. A single coin toss cannot give both a head and a tail. A person cannot run forwards and backwards at the same instant.
Note 10.11. Mutually exclusive is not the same as independent, though the words feel similar. Two mutually exclusive events are as far from independent as possible: knowing that one has happened tells you the other certainly has not.
10.3.1 Rules of probability
1. Complement.
Definition 10.12. The complement of \(E\), written \(E^c\), is the event that \(E\) does not happen, and \[P\left (E^c\right )=1-P(E).\]
Note 10.13. This holds because \(E\) and \(E^c\) between them cover the whole sample space and share nothing, so their probabilities must add to \(1\). It is worth reaching for whenever a question asks for “at least one” — counting the ways of getting none and subtracting is usually far shorter than counting all the ways of getting some.
2. Addition rule.
Definition 10.14. For any two events \(E_1\) and \(E_2\), \[P\left (E_1\cup E_2\right )=P\left (E_1\right )+P\left (E_2\right )-P\left (E_1\cap E_2\right ),\] which reduces to \[P\left (E_1\cup E_2\right )=P\left (E_1\right )+P\left (E_2\right )\] when \(E_1\) and \(E_2\) are mutually exclusive.
Note 10.15. The subtraction is there because outcomes belonging to both events get counted once in \(P(E_1)\) and again in \(P(E_2)\). Taking away \(P(E_1\cap E_2)\) removes the duplication. When the events are mutually exclusive there is no overlap to remove, and the term is zero.
Example 10.16. Two dice are thrown. Find the probability of scoring either a double or a total less than \(5\).
Solution. Let \(E_1\) be the event of a double and \(E_2\) the event of a total less than \(5\). Listing them: \[E_1=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\},\qquad n\left (E_1\right )=6,\] \[E_2=\{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\},\qquad n\left (E_2\right )=6.\] The outcomes in both lists are \((1,1)\) and \((2,2)\), so \(n\left (E_1\cap E_2\right )=2\). The word “or” calls for the addition rule, and with \(n(S)=36\): \[P\left (E_1\cup E_2\right )=\frac {6}{36}+\frac {6}{36}-\frac {2}{36}=\frac {10}{36} =\frac {5}{18}.\]
Note 10.17. Adding \(\frac {6}{36}+\frac {6}{36}=\frac {1}{3}\) and stopping counts \((1,1)\) and \((2,2)\) twice each. The events are not mutually exclusive — a double can give a total below \(5\) — so the overlap has to come out.
3. Conditional probability.
Definition 10.18. The probability that \(E_2\) happens given that \(E_1\) has already happened is \[P\left (E_2\mid E_1\right )=\frac {P\left (E_2\cap E_1\right )}{P\left (E_1\right )}.\]
Note 10.19. Some texts, including the departmental notes, write this as \(P(E_2/E_1)\). The vertical bar and the slash mean exactly the same thing; the bar is used here because a slash is easily misread as division.
Note 10.20. Being told that \(E_1\) happened shrinks the sample space: outcomes outside \(E_1\) are no longer possible. Dividing by \(P(E_1)\) is what rescales the smaller space so that its probabilities still add to \(1\).
Example 10.21. Two dice are thrown. Find the probability of scoring a double given that the total is less than \(5\).
Solution. Using the same two events as before, \[P\left (E_1\cap E_2\right )=\frac {2}{36},\qquad P\left (E_2\right )=\frac {6}{36}.\] \[P\left (E_1\mid E_2\right )=\frac {P\left (E_1\cap E_2\right )}{P\left (E_2\right )} =\frac {\frac {2}{36}}{\frac {6}{36}}=\frac {2}{6}=\frac {1}{3}.\]
Without the condition, a double has probability \(\frac {6}{36}=\frac {1}{6}\). Knowing the total is below \(5\) doubles it, because low totals are made from few combinations and doubles form a larger share of them.
Note 10.22. The order matters. \(P\left (E_1\mid E_2\right )=\frac {1}{3}\) is the chance of a double given a low total; \(P\left (E_2\mid E_1\right )=\frac {2}{6}=\frac {1}{3}\) happens to agree here only because the two events have the same size. In general the two are different questions with different answers, and reading the sentence carefully to see which is being asked is half the work.
Rearranging the definition gives a rule for “and” to sit beside the rule for “or”: \[P\left (E_1\cap E_2\right )=P\left (E_2\mid E_1\right )P\left (E_1\right ).\]
Example 10.23. A crate holds \(8\) mangoes, of which \(3\) are bad. Two are taken at random, one after the other, without replacement. Find the probability that both are bad.
Solution. Let \(E_1\) be the event that the first mango is bad and \(E_2\) that the second is bad. We want \(P\left (E_1\cap E_2\right )\).
First pick. Three of the eight are bad: \[P\left (E_1\right )=\frac {3}{8}.\]
Second pick. A bad mango has been removed and not replaced, so only \(7\) remain and only \(2\) of them are bad: \[P\left (E_2\mid E_1\right )=\frac {2}{7}.\]
Multiply. \[P\left (E_1\cap E_2\right )=P\left (E_2\mid E_1\right )P\left (E_1\right ) =\frac {2}{7}\times \frac {3}{8}=\frac {6}{56}=\frac {3}{28}.\]
Note 10.24. Both the numerator and the denominator change on the second pick. Reducing the total from \(8\) to \(7\) but leaving the count of bad ones at \(3\) gives \(\frac {3}{7}\) and a wrong answer — the mango that was removed was itself a bad one.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.