Mutually Exclusive Events
Mutually exclusive events are two events that cannot occur simultaneously. In short, if one event occurs, then the other cannot at the same time.
For example, let’s say you roll a die. If you roll a three, then you cannot roll a one, two, four, five, nor a six at the same time. In this case, rolling a three is mutually exclusive to rolling any other number.
When represented mathematically, if two events, A and B, are mutually exclusive, then:
\(P(A \cap B) = 0\)
as there is no probability of these two events occurring at the same time.
Given two independent events, \(A\) and \(B\), we use \(A \cup B\) to denote either events \(A\), \(B\), or both happening.
To calculate the probability of two mutually exclusive events, \(A\) or \(B\) to occur, we can use the following formula:
\(P(A \cup B) = P(A) + P(B)\)
ExampleYou roll a six-sided die. What is the probability that you either roll a six or roll an odd number?
First, we can let \(6\) denote the event of rolling a six and \(O\) denote the event of rolling an odd number.
Since rolling a six and rolling an odd number are mutually exclusive, we can use the aforementioned formula:
\(P(6 \cup O) = P(6) + P(O)\)
\(P(6 \cup O) = \cfrac{1}{6} + \cfrac{3}{6}\)
\(P(6 \cup O) = \cfrac{4}{6} = 0.6667\)
Therefore, the probability of rolling a six or an odd number is \(\boldsymbol{0.6667}\) (or \(\boldsymbol{66.67\%}\)).
In a regular deck of \(52\) cards, you draw either a black card or a heart card.
Drawing a black card or a heart card are mutually exclusive events.
First, we can let \(B\) denote drawing a black card and \(H\) denote drawing a heart card.
Next, we can determine the probability of drawing either a black card or heart card as follows:
\(P(B \cup H) = P(B) + P(H)\)
\(P(B \cup H) = \cfrac{26}{52} + \cfrac{13}{52}\)
\(P(B \cup H) = \cfrac{39}{52} = \cfrac{3}{4} = 0.75\)
Therefore, we can determine the probability of drawing either a black or heart card is \(\boldsymbol{0.75}\) (or \(\boldsymbol{75\%}\)).
You toss two coins. You either get one head and one tail or you get at least one tail.
Getting both a head and tail and getting at least one tail are not mutually exclusive events.
You roll two dice. You either roll a value less than or equal to \(5\) or you roll a \(9\).
Getting a combined value of \(5\) or less or getting a \(9\) are mutually exclusive events.
First, we can let \(X\) denote the rolled number.
We have \(36\) possible combinations of numbers with two dice. This will be our denominator when calculating the probabilities of these different events.
The possible dice combinations to get a \(5\) are \((1, 4)\), \((2, 3)\), \((3, 2)\), \((4, 1)\).
The possible dice combinations to get a \(4\) are \((1, 3)\), \((2, 2)\), \((3, 1)\).
The possible dice combinations to get a \(3\) are \((1, 2)\), \((2, 1)\).
The possible dice combinations to get a \(2\) are \((1, 1)\).
The calculation of the probability of rolling a number less than or equal to \(5\) is as follows:
\(P(X \le 5) = P(X=2) + P(X=3) + P(X=4) + P(X=5)\)
\(P(X \le 5) = \cfrac{4}{36} + \cfrac{3}{36} + \cfrac{2}{36} + \cfrac{1}{36}\)
\(P(X \le 5) = \cfrac{10}{36}\)
The possible dice combinations to get a \(9\) are \((6, 3)\), \((5, 4)\), \((4, 5)\), \((6, 3)\).
Therefore, the probability to rolling a \(9\) is \(\cfrac{4}{36}\).
Now we can calculate the probability of rolling a number less than or equal to \(5\) or rolling a \(9\) as such:
\(P(X \le 5 \cup X = 9) = P(X \le 5) + P(X = 9)\)
\(P(X \le 5 \cup X = 9) = \cfrac{10}{36} + \cfrac{4}{36}\)
\(P(X \le 5 \cup X = 9) = \cfrac{14}{36} = \cfrac{7}{18} = 0.3888\)
Therefore, we can determine the probability of rolling either a total value less than or equal to \(5\) or rolling a \(9\) with \(2\) dice is \(\boldsymbol{0.3889}\) (or \(\boldsymbol{0.3889}\)).