Curriculum Alignment
- Strand
- 3.0 Statistics and Probability
- Sub-Strand
- 3.2 Probability 1
- Specific Learning Outcomes
Subsection 3.2.4 Independent Events
Teacher Resource 3.2.20.
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 3.2.5.
\({\color{black} \textbf{Work in groups}}\)
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Define Independent events
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State one example of two events that are independent
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A student can choose to join either the Science Club or the Drama Club, but not both.
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If the probability of joining Science Club is \(40\%\) and Drama Club is \(30\%\) , what is the probability that a student joins either club?
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Are these events mutually exclusive or independent? Explain.
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Compare and discuss answers with other groups
\({\color{black} \textbf{Key Takeaway}}\)
Two events are independent if the occurrence of one does not affect the probability of the other occurring.
If \(A\) and \(B\) are independent events, then;
\begin{gather*}
P(\textbf{A and B}) = \textbf{P(A)} \times \textbf{P(B)}
\end{gather*}
\begin{gather*}
\textbf{P}(\textbf{A} \cap \textbf{B}) \, = \, \textbf{P(A)} \times \textbf{P(B)}
\end{gather*}
This means the probability of both events occurring together is the product of their individual probabilities.
\(\text{Example}\)
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A student bringing a lunch from home and another student buying a lunch from the cafeteria.
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A student answering a question correctly in english class and another student dropping their pencil in science class.
Example 3.2.21.
A coin is flipped, and a six-sided die is rolled. What is the probability of getting heads and rolling a 6?
Solution.
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the Sample Space isPossible coin outcomes \(\textbf{{H, T}}\)Possible die outcomes \(\textbf{{1, 2, 3, 4, 5, 6}}\)
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Favorable Outcomes\(\textbf{P(H)} = \frac{1}{2}\)\(\textbf{P(6)} = \frac{1}{6}\)
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Since flipping the coin and rolling the die are independent events\begin{gather*} P(\textbf{H and 6}) = \textbf{P(H)} \times \textbf{P(6)} \end{gather*}\begin{gather*} = \frac{1}{2} \times \frac{1}{6} \end{gather*}\begin{gather*} = \frac{1}{12} \end{gather*}
Example 3.2.22.
The probability that it rains on a given day is \(40\%\text{,}\) and the probability that a person is late to work is \(20\%\text{.}\)
Let \(\textbf{P(R)}\) represent the probability that it rains and \(\textbf{P(L)}\) be the probability that the person is late. The compliment of an event is the probability that it does not happen. Therefore, \(\textbf{P}(\textbf{R}^{c})\) will represent the probability that it does not rain and \(\textbf{P}(\textbf{L}^{c})\) the probability that the person is not late.
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Find the probability that it rains \(\textbf{P(R)}\) .
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Find the probability that the person is late \(\textbf{P(L)}\text{.}\)
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Find the probability that it rains and the person is late.
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Find the probability that it does not rain but the person is late.
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Find the probability that it rains or the person is late.
Solution.
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The probability that it rains is\begin{gather*} \textbf{P(R) = 0.4} \end{gather*}The probability that it rains is \(\textbf{0.4 (or 40\%)}\text{.}\)
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The probability that the person is late is\begin{align*} \textbf{P(L) = 0.2} \amp \end{align*}
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Since rain and being late are independent,\begin{gather*} \textbf{P}(\textbf{R} \, \cap \textbf{L}) \, = \, \textbf{P(R)} \times \textbf{P(L)} \end{gather*}\begin{gather*} \textbf{P}(\textbf{R} \, \cap \textbf{L}) \, = \, 0.4 \times 0.2 \, = \, 0.08 \end{gather*}The probability that it rains and the person is late is \(\textbf{0.08 (or 8\%)}\text{.}\)
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The probability that it does not rain is\begin{gather*} \textbf{P}(\textbf{R}^{c}) \, = \, \textbf{1 - P(R)} = 1 \, - \, 0.4 \, = \, 0.6 \end{gather*}Since rain and being late are independent\begin{gather*} \textbf{P}(\textbf{R}^{c} \cap L) = \textbf{P}(\textbf{R}^{c}) \times \textbf{P(L)} \end{gather*}\begin{gather*} \textbf{P}(\textbf{R}^{c} \cap L) = 0.6 \, \times \, 0.2 \, = \, 012 \end{gather*}
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The probability that it rains or the person is late is\begin{gather*} \textbf{P}(\textbf{R} \cup \textbf{L}) = \textbf{P(R)} \, + \, \textbf{P(R)} \, - \, \textbf{P}(\textbf{R} \cup \textbf{L}) \end{gather*}\begin{gather*} \textbf{P}(\textbf{R} \cup \textbf{L}) = 0.4 \, + \, 0.2 \, - \, 0.08 \end{gather*}\begin{gather*} = \, 0.52 \end{gather*}
Checkpoint 3.2.23.
Exercises Exercises
1.
A coin is tossed twice. What is the probability of getting heads on both tosses?
2.
A die is rolled, and a coin is tossed. What is the probability of rolling a 6 on the die and getting tails on the coin?
3.
A bakery in Makongeni produces cakes. The probability that a cake is decorated with chocolate icing is \(0.7\text{.}\) If two cakes are made independently, what is the probability that both cakes are decorated with chocolate icing?
4.
A seed has a \(60\%\) chance of germinating. If two seeds are planted independently, what is the probability that both seeds germinate?
