Curriculum Alignment
- Strand
- 3.0 Statistics and Probability
- Sub-Strand
- 3.2 Probability 1
- Specific Learning Outcomes
- Determine the probability of mutually exclusive and independent events
Subsection 3.2.3 Mutually Exclusive Events
Teacher Resource 3.2.15.
To assist your teaching, we have prepared lesson resources, aligned with this textbook and the CBC. The Lesson Plan links to syllabus learning outcomes and provides suggest time allocations. The Step-by-Step Guide provides more detailed guidance on how to teach the content, including suggested questions to ask learners, and possible answers.
Learner Experience 3.2.4.
\({\color{black} \textbf{Work in groups}}\)
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Define Mutually exclusive events
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State one example of Two events that are mutually exclusive
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In a class of 40 students, 18 take French, 22 take German, and no student takes both.
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What is the probability that a randomly selected student takes French or German?
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Are the events “taking French” and “taking German” mutually exclusive? Explain.
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Compare and discuss answers with other groups
\({\color{black} \textbf{Key Takeaway}}\)
Two events are mutually exclusive if they cannot occur at the same time.
This means that if one event happens, the other cannot.
If \(\textbf{A} \) and \(\text{B}\) are mutually exclusive events, then;
\begin{gather*}
\textbf{P(A and B) = 0}
\end{gather*}
\begin{gather*}
\textbf{P}(\textbf{A} \cap \textbf{B} ) \, = \, 0
\end{gather*}
The probability of either \(\textbf{A}\) or \(\textbf{B}\) occurring is;
\begin{gather*}
\textbf{P(A or B) = P(A) + P(B)}
\end{gather*}
\(\text{Example} \)
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A traffic light being red and the same traffic light being green at the exact same time.
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You are sleeping and you are wide awake at the exact same moment.
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A door being open and the same door being closed at the same time.
Example 3.2.16.
Solution.
Favorable Outcomes
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\(\displaystyle \textbf{P(3)} = \frac{1}{6}\)
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\(\displaystyle \textbf{P(5)} = \frac{1}{6}\)
Since rolling a 3 and rolling a 5 are mutually exclusive events;
\begin{gather*}
\textbf{P(3 or 5)} = \textbf{P(3) + P(5)}
\end{gather*}
\begin{gather*}
= \frac{1}{6} + \frac{1}{6}
\end{gather*}
\begin{gather*}
= \frac{2}{6} = \frac{1}{3}
\end{gather*}
Example 3.2.17.
A card is drawn from a standard deck of 52 playing cards. Let Event A be drawing a Heart and Event B be drawing a Spade. Are these events mutually exclusive? Explain
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Find \(\textbf{P(A) and P(B)} \)
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Calculate \(\textbf{P(A or B)}\)
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What is \(\textbf{P}(\textbf{A} \, \cap \, \textbf{B})\)
Solution.
Two events are mutually exclusive if they cannot happen at the same time. Since a card cannot be both a Heart and a Spade, the events are mutually exclusive.
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We calculate the probability of drawing a Heart or a Spade.Since there are \(\textbf{13 Hearts}\) in the deck,\begin{gather*} \textbf{P(A)} \, = \, \frac{13}{52} \end{gather*}Since there are \(\textbf{13 Spades}\) in the deck,\begin{gather*} \textbf{P(B)} \, = \, \frac{13}{52} \end{gather*}
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For mutually exclusive events, we use,\begin{gather*} \textbf{P(A or B) = P(A) + P(B)} \end{gather*}\begin{gather*} = \frac{13}{52} + \frac{13}{52} \end{gather*}\begin{gather*} = \frac{26}{52} \end{gather*}\begin{gather*} = \frac{1}{2} \end{gather*}
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Since these events are mutually exclusive, their intersection is zero:\begin{gather*} \textbf{P} (\textbf{A} \cap \textbf{B}) = 0 \end{gather*}
Checkpoint 3.2.18.
Checkpoint 3.2.19.
Exercises Exercises
1.
A student at Khungu Senior Secondary School tosses a coin once. Are the events "getting heads" and "getting tails" mutually exclusive?
Answer.
2.
A person selects one piece of fruit from a bowl containing apples, bananas, and oranges.Is selecting an apple and selecting a banana mutually exclusive events?
Answer.
Yes, selecting an apple and selecting a banana are mutually exclusive events because they cannot happen at the same time. When you select one piece of fruit from the bowl, you can only choose one type of fruit. Therefore, if you select an apple, you cannot select a banana, and if you select a banana, you cannot select an apple.
3.
A student recorded their method of transport to school for 10 days. The methods used were:
\begin{gather*}
\textbf{Walk, Bus, Bus, Walk, Bike, Walk, Bus, Bike, Walk, Bus}
\end{gather*}
Let Event A = "The student walked to school" and Event B = "The student took the bus to school"
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Are events A and B mutually exclusive? Explain your answer.
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What is the probability that the student either walked or took the bus to school on a randomly chosen day?
Answer.
(a) Yes, events A and B are mutually exclusive because they cannot happen at the same time. The student cannot walk to school and take the bus on the same day. Therefore, if event A occurs (the student walked), event B cannot occur (the student took the bus), and vice versa.
(b) To find the probability that the student either walked or took the bus to school on a randomly chosen day, we first count the number of days the student walked and the number of days they took the bus. The student walked on 5 days and took the bus on 5 days. Since these events are mutually exclusive, we can add their probabilities: \[ P(A \text{ or } B) = P(A) + P(B) \] The probability of walking is \( P(A) = \frac{5}{10} = \frac{1}{2} \) and the probability of taking the bus is \( P(B) = \frac{5}{10} = \frac{1}{2} \). Therefore, \[ P(A \text{ or } B) = \frac{1}{2} + \frac{1}{2} = 1 \] So, the probability that the student either walked or took the bus to school on a randomly chosen day is 1, meaning it is certain that they did one of those two things.
4.
A card is drawn from a standard 52-card deck.
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Are the events "drawing a diamond" and "drawing a club" mutually exclusive?
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What is the probability of drawing either a diamond or a club?
Answer.
(a) Yes, the events "drawing a diamond" and "drawing a club" are mutually exclusive because they cannot happen at the same time. A card cannot be both a diamond and a club simultaneously. Therefore, if you draw a diamond, you cannot draw a club, and if you draw a club, you cannot draw a diamond.
(b) To find the probability of drawing either a diamond or a club, we first determine the number of diamonds and clubs in a standard deck of 52 cards. There are 13 diamonds and 13 clubs in the deck. Since these events are mutually exclusive, we can add their probabilities: \[ P(\text{diamond or club}) = P(\text{diamond}) + P(\text{club}) \] The probability of drawing a diamond is \( P(\text{diamond}) = \frac{13}{52} = \frac{1}{4} \) and the probability of drawing a club is \( P(\text{club}) = \frac{13}{52} = \frac{1}{4} \). Therefore, \[ P(\text{diamond or club}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \] So, the probability of drawing either a diamond or a club is \( \frac{1}{2} \) or 50%.
5.
In a lottery competition, there are five cards labelled A, B, C, D, and F. A player must pick only one card to enter the competition.
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Are the events "picking card A" and "picking card F" mutually exclusive?
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What is the probability of either picking card A or picking card F?
Answer.
(a) Yes, the events "picking card A" and "picking card F" are mutually exclusive because they cannot happen at the same time. A player can only pick one card, so if they pick card A, they cannot pick card F, and if they pick card F, they cannot pick card A.
(b) To find the probability of either picking card A or picking card F, we first determine the total number of cards, which is 5. Since there is only one card A and one card F, the probability of picking card A is \( P(A) = \frac{1}{5} \) and the probability of picking card F is \( P(F) = \frac{1}{5} \). Since these events are mutually exclusive, we can add their probabilities: \[ P(A \text{ or } F) = P(A) + P(F) \] Therefore, \[ P(A \text{ or } F) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5} \] So, the probability of either picking card A or picking card F is \( \frac{2}{5} \) or 40%.
