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 eventsĀ \(A\ \)and\(\ B\), that is\(\ P(A\ or\ B) =\)Ā \(P(A \cup B)\)Ā isĀ \(P(A) + P(B)\)
Example:Ā 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:Ā If eventĀ \(A\)Ā = obtaining a prime number and eventĀ \(B\)Ā = obtaining an odd number in a single throw of dice. Find the probability of eventsĀ \(A\)Ā andĀ \(B\)
Solution
\[A = \left\{ 2,3,5 \right\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = \{ 1,3,5\}\]
\[P(A) = \frac{3}{6} = \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ P(B) = \frac{3}{6} = \frac{1}{2}\]
\[A \cap B = \{ 3,5\}\]
\[P(A \cap B) = \frac{2}{6} = \frac{1}{3}\]
\[P(A \cup B) = P(A) + P(B) - P(A \cap B) = \ \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = \ \frac{2}{3}\]