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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 3.2.5 Laws of Probability

The laws of probability are fundamental principles that guide us in calculating and understanding the chances of events occurring. These rules provide a systematic way to solve probability problems in mathematics and to analyze outcomes in random experiments.
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Important laws such as the addition rule, multiplication rule, and complement rule help us determine the likelihood of single or combined events and compute probabilities accurately.
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Subsubsection 3.2.5.1 Addition Rule

Learner Experience 3.2.6.

Work in groups
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(a)
A class has \(40\) students. \(22\) students play Football, \(18\) students play Volleyball, and \(8\) students play both Football and Volleyball.
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  1. Find the probability that a randomly selected student plays Football.
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  2. Find the probability that a randomly selected student plays Volleyball.
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  3. If you add the two probabilities from (a) and (b), what do you notice?
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Hint.
Remember that some students are counted in both groups.
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(b)
Find the probability that a randomly selected student plays Football or Volleyball.
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  1. Compare your answer with the sum of the probabilities in part (a)(\(3\text{.}\)). Why is there a difference?
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  2. Write a general formula for finding the probability of Event \(A\) or Event \(B\text{.}\)
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Key Takeaway 3.2.25.

The addition rule of probability helps us find the probability that at least one of two events occurs. If A and B are two events, the probability that event A or event B (or both) occurs is given by:
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\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
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Here, \(P(A \cap B)\) represents the probability that both events A and B occur at the same time.
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We subtract \(P(A \cap B)\) because the outcomes that belong to both events would otherwise be counted twice.
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This rule is used when events A and B are not mutually exclusive, meaning they can occur together.
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Example 3.2.26.

A standard deck of cards contains \(52\) cards. There are \(13\) hearts and \(13\) clubs in the deck. If one card is drawn at random, find the probability that the card is a heart or a club.
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Solution.
Since a single card cannot be both a heart and a club at the same time, the two events are mutually exclusive.
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First, find the probability of each event:
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  • \(\displaystyle P(Heart) = \frac{13}{52}\)
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  • \(\displaystyle P(Club) = \frac{13}{52}\)
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Because the events are mutually exclusive, we apply the addition rule:
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\begin{align} P(Heart or Club) \amp = P(Heart) + P(Club) \tag{3.2.1}\\ \amp = \frac{13}{52} + \frac{13}{52} \tag{3.2.2}\\ \amp = \frac{26}{52} = \frac{1}{2}\tag{3.2.3} \end{align}
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Therefore, the probability of drawing either a heart or a club is \(\frac{1}{2}\text{,}\) which is equal to \(0.5\) or \(50\%\text{.}\)
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Example 3.2.27.

You roll a fair six-sided die. What is the probability of rolling a number divisible by \(2\) or \(3\text{?}\)
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Solution.
A fair six-sided die has \(6\) equally likely outcomes: \(\{1,2,3,4,5,6\}\text{.}\)
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List the numbers divisible by \(2\text{:}\)
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\(\{2,4,6\}\) → \(3\) numbers
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List the numbers divisible by \(3\text{:}\)
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\(\{3,6\}\) → \(2\) numbers
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We want the numbers divisible by \(2\) or \(3\) (the union of the two sets).
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Notice that \(6\) appears in both sets, so we must avoid counting it twice.
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Using the inclusion–exclusion principle:
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\begin{gather*} P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \end{gather*}
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Compute each probability:
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  • \(\displaystyle P(\text{divisible by 2}) = \frac{3}{6}\)
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  • \(\displaystyle P(\text{divisible by 3}) = \frac{2}{6}\)
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  • \(\displaystyle P(\text{divisible by both 2 and 3}) = P(\text{divisible by 6}) = \frac{1}{6}\)
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\begin{gather*} P(\text{divisible by 2 or 3}) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \end{gather*}
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Therefore, the probability of rolling a number divisible by \(2\) or \(3\) is \(\frac{2}{3}\text{.}\)
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Checkpoint 3.2.28.

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Checkpoint 3.2.29.

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Checkpoint 3.2.30.

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Exercises Exercises

1.
A student can get an A, B, C, D, or F in a class. What is the probability that the student gets an A or a B?
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Answer.
\(\frac{2}{5}\) or \(40\%\text{.}\)
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2.
A die is rolled. What is the probability of rolling a 1 or a 6?
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Answer.
\(\frac{1}{3}\) or \(33.3\%\text{.}\)
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3.
In a class of 30 students, 15 students like math, 10 students like chemistry, and 5 students like both math and chemistry. What is the probability that a randomly chosen student likes math or chemistry?
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Answer.
\(\frac{5}{6}\) or \(83.3\%\text{.}\)
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4.
A bag contains 8 blue marbles and 5 yellow marbles. What is the probability of drawing a blue marble or a yellow marble?
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Answer.
\(1\) or \(100\%\text{.}\)
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5.
In a class of 25 students, 12 play soccer, 10 play basketball, and 5 play both. What is the probability that a randomly chosen student plays soccer or basketball?
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Answer.
\(\frac{17}{25}\) or \(68\%\text{.}\)
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6.
A bag contains letters of the word \(MATHEMATICS\text{.}\) What is the probability of selecting a vowel or the letter \(M\text{?}\)
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Answer.
\(\frac{7}{13}\) or \(53.8\%\text{.}\)
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7.
A number is chosen between 1 and 10. What is the probability that it is a 3 or a 7?
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Answer.
\(\frac{1}{5}\) or \(20\%\text{.}\)
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8.
A day of the week is chosen at random. What is the probability that it is a Saturday or a Sunday?
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Answer.
\(\frac{2}{7}\) or \(28.6\%\text{.}\)
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Subsubsection 3.2.5.2 Multiplication Rule

Learner Experience 3.2.7.

\({\color{black} \textbf{Work in groups}}\)
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A factory produces \(90\%\) good items and \(10\%\) defective items. A quality check is performed on two randomly selected items
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  1. Find the probability that both items are good.
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  2. Find the probability that at least one item is defective.
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  3. Are these events independent? Explain.
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\({\color{black} \textbf{Key Takeaway}}\)
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The multiplication rule is used to find the probability of two events happening together.
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For \(\text{Independent events}\text{,}\)the probability of both occurring is
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\begin{gather*} \textbf{P(A and B)} = \textbf{P(A)} \times \textbf{P(B)} \end{gather*}
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For \(\text{Dependent events}\)
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\begin{gather*} \textbf{P}(\textbf{A} \cap \textbf{B} ) \, = \, \textbf{P(A)} \, \times \, \textbf{P(B|A)} \end{gather*}
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Here, \(\textbf{P(B|A)}\) is the probability that B happens given that A has already occurred.
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Example 3.2.32.

A student in Modegashe primary school was instructed to roll a Die and Toss a Coin.
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What was the probability of rolling a 4 on the die and getting head on the coin?
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Solution.
  1. Probability of rolling a 4 on a six-sided die is
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    \begin{gather*} \frac{1}{6} \end{gather*}
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  2. Probability of getting heads on the coin is
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    \begin{gather*} = \frac{1}{2} \end{gather*}
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\begin{gather*} \textbf{so, } \, \textbf{P(4 and H)} = \frac{1}{6} \times \frac{1}{2} \end{gather*}
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\begin{gather*} = \frac{1}{12} \end{gather*}
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the probability of rolling a 4 and flipping heads is \(\frac{1}{12}\) or \(8.33\%\)
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Example 3.2.33.

A person has a \(60\%\) probability of catching the first bus and an \(80\%\) probability of catching the second bus (if they miss the first one)
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  1. Find the probability that the person catches the first bus.
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  2. Find the probability that the person misses the first bus but catches the second.
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  3. Find the probability that the person misses both buses.
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Solution.
Probability of catching the first bus:
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\begin{gather*} \textbf{P(A) = 0.6} \end{gather*}
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Probability of missing the first bus
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\begin{gather*} \textbf{P}(\textbf{A}^{c}) \, = \, 1 - \textbf{P(A)} \, = \, 1 - 0.6 \, = \, 0.4 \end{gather*}
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Probability of catching the second bus, given that the first bus was missed
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\begin{gather*} \textbf{P}(\textbf{B}|\textbf{A}^{c}) \, = \, 0.8 \end{gather*}
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Probability of missing the second bus, given that the first bus was missed
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\begin{gather*} \textbf{P}(\textbf{B}^{c} | \textbf{A}^{c} ) \, = \, 1 - \textbf{P}(\textbf{B} | \textbf{A}^{c} ) \, = \, 1 - 0.8 \, = \, 0.2 \end{gather*}
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  1. The probability of catching the first bus is directly given as
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    \begin{gather*} \textbf{P(A)} \, = \, 0.6 \end{gather*}
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    The probability of catching the first bus is\(0.6\) or \(60\%\text{.}\)
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  2. Missing the first bus \(\textbf{A}^{c}\)
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    Catching the second bus \(\text{B}\)
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    Since these events are dependent, we use the multiplication rule
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    \begin{gather*} \textbf{P}(\textbf{A}^{c} \cap \textbf{B} ) \, = \, \textbf{P}(\textbf{A}^{c}) \times \textbf{P} (\textbf{B} | \textbf{A}^{c}) \end{gather*}
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    \begin{gather*} \textbf{P}(\textbf{A}^{c} \cap \textbf{B} ) \, = \, 0.4 \times 0.8 \, = \, 0.32 \end{gather*}
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    The probability of missing the first bus but catching the second is \(0.32\) or \(32\%\text{.}\)
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  3. Missing the first bus \(\textbf{A}^{c}\)
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    Missing the second bus \(\textbf{B}^{c}\)
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    Again, using the multiplication rule
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    \begin{gather*} \textbf{P}(\textbf{A}^{c} \cap \textbf{B}^{c} ) \, = \, \textbf{P(A)}^{c} \times \textbf{P}(\textbf{B}^{c}|\textbf{A}^{c}) \end{gather*}
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    \begin{gather*} \textbf{P}(\textbf{A}^{c} \cap \textbf{B}^{c} ) \, = \, 0.4 \times 0.2 \, = \, 0.08 \end{gather*}
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    The probability of missing both buses is \(0.08\) or \(8\%\text{.}\)
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Checkpoint 3.2.34. Multiplication Rule.

Load the question by clicking in the button below.
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Exercises Exercises

1.
A coin is tossed twice. What is the probability of getting heads on both tosses?
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Answer.
\(\frac{1}{4}\) or \(25\%\text{.}\)
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2.
A die is rolled, and a coin is tossed. What is the probability of rolling a 6 and getting tails?
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Answer.
\(\frac{1}{12}\) or \(8.33\%\text{.}\)
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3.
A weather forecast predicts a \(60\%\) chance of sunshine on Monday and a \(70\%\) chance of sunshine on Tuesday. Assuming these forecasts are independent, what is the probability of sunshine on both Monday and Tuesday?
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Answer.
\(0.42\) or \(42\%\text{.}\)
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4.
A farmer in Gakuonyo plants two seeds. Each seed has a \(75\%\) chance of germinating. What is the probability that both seeds germinate?
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Answer.
\(0.5625\) or \(56.25\%\text{.}\)
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