Addition of Probabilities - SS2 Mathematics Lesson Note

The probability of the occurrence of any two (or more) mutually exclusive events is equal to the sum of their individual probabilities.

The probability of two mutually exclusive events \(A\ \)and\(\ B\), that is\(\ P(A\ or\ B) =\) \(P(A \cup B)\) is \(P(A) + P(B)\)

Example: A bag contains 2 red balls, 4 blue balls and 6 green balls. What is the probability of drawing a red or blue ball?

Solution

\[total\ number\ of\ balls = 2 + 4 + 6 = 12\]

\[P(r) = \ \frac{2}{12},\ \ P(b) = \ \frac{4}{12},\ \ P(g) = \ \frac{6}{12}\]

\[P(r\ or\ b) = P(r) + P(b) = \frac{2}{12} + \frac{4}{12} = \frac{6}{12} = \frac{1}{2}\]

The probability of the occurrence of any two (or more) non-mutually exclusive events is equal to the subtraction of their common outcomes from the sum of their individual probabilities. Mathematically,

\[P(A\ or\ B) = P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Example: If event \(A\) = obtaining a prime number and event \(B\) = obtaining an odd number in a single throw of dice. Find the probability of events \(A\) and \(B\)

Solution

\[A = \left\{ 2,3,5 \right\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \{ 1,3,5\}\]

\[P(A) = \frac{3}{6} = \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ P(B) = \frac{3}{6} = \frac{1}{2}\]

\[A \cap B = \{ 3,5\}\]

\[P(A \cap B) = \frac{2}{6} = \frac{1}{3}\]

\[P(A \cup B) = P(A) + P(B) - P(A \cap B) = \ \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \ \frac{2}{3}\]