Subsubsection 3.2.5.1 Addition Law of Probability
Activity 3.2.6.
\({\color{black} \textbf{Work in groups}}\)
The probability that a student passes Mathematics is \(75\%\) and the probability that they pass English is \(60\%\text{.}\) If the probability of passing both is \(50\%\text{,}\) find the probability that the student passes either Mathematics or English.
compare answers with other groups
\({\color{black} \textbf{Key Takeaway}}\)
\(\text{The addition law}\) is used to find the probability of either one event or another occurring.
\(\text{Mutually exclusive events}\)
The probability of either event occurring is;
\begin{gather*}
\textbf{P}(\textbf{A} \cup \textbf{B} ) \, = \, \textbf{P(A) + P(B)}
\end{gather*}
\begin{gather*}
\textbf{P(A or B) = P(A) + P(B)}
\end{gather*}
Since mutually exclusive events cannot happen at the same time, \(\textbf{P}(\textbf{A} \cap \textbf{B} ) \, = \, 0\)
\(\text{Non-mutually exclusive events}\)
If two events can happen at the same time, we must subtract the probability of them happening together to avoid double counting.
That is;
\begin{gather*}
\textbf{P}(\textbf{A} \cup \textbf{B} ) \, = \, \textbf{P(A) + P(B)} \, - \, \textbf{P}(\textbf{A} \cap \textbf{B} )
\end{gather*}
\begin{gather*}
\textbf{P(A or B) = P(A) + P(B) - P(A and B)}
\end{gather*}
Example 3.2.9.
A standard deck has \(52\) cards, with \(13\) hearts and \(13\) clubs.
Since a single card cannot be both a heart and a club, the events are mutually exclusive.
Solution.
\begin{gather*}
P(\textbf{Heart or Club}) = \frac{13}{52} + \frac{13}{52}
\end{gather*}
\begin{gather*}
= \frac{26}{52} = \frac{1}{2}
\end{gather*}
the probability of drawing either a heart or a club is \(0.5\) or \(50\%\)
Example 3.2.10.
now,
\begin{gather*}
P(\textbf{Red}) = \frac{26}{52}
\end{gather*}
\begin{gather*}
P(\textbf{Heart}) = \frac{13}{52}
\end{gather*}
\begin{gather*}
P(\textbf{Red and Heart}) = \frac{13}{52}
\end{gather*}
\begin{gather*}
P(\textbf{Red or Heart}) = \frac{26}{52} + \frac{13}{52} - \frac{13}{52} = \frac{26}{52}
\end{gather*}
\begin{gather*}
= \frac{1}{2}
\end{gather*}
the probability of drawing either a red card or a king is \(0.50\) or \(50\%\)
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?
2.
A die is rolled. What is the probability of rolling a 1 or a 6?
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?
4.
A bag contains 8 blue marbles and 5 yellow marbles. What is the probability of drawing a blue marble or a yellow marble?
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?
6.
A bag contains letters of the word \(MATHEMATICS\text{.}\) What is the probability of selecting a vowel or the letter \(M\text{?}\)
7.
A number is chosen between 1 and 10. What is the probability that it is a 3 or a 7?
8.
A day of the week is chosen at random. What is the probability that it is a Saturday or a Sunday?
Subsubsection 3.2.5.2 multiplication rule
Activity 3.2.7.
\({\color{black} \textbf{Work in groups}}\)
A factory produces \(90\%\) good items and \(10\%\) defective items. A quality check is performed on two randomly selected items
Find the probability that both items are good.
Find the probability that at least one item is defective.
Are these events independent? Explain.
\({\color{black} \textbf{Key Takeaway}}\)
The multiplication rule is used to find the probability of two events happening together.
For \(\text{Independent events}\text{,}\)the probability of both occurring is
\begin{gather*}
\textbf{P(A and B)} = \textbf{P(A)} \times \textbf{P(B)}
\end{gather*}
For \(\text{Dependent events}\)
\begin{gather*}
\textbf{P}(\textbf{A} \cap \textbf{B} ) \, = \, \textbf{P(A)} \, \times \, \textbf{P(B|A)}
\end{gather*}
Here, \(\textbf{P(B|A)}\) is the probability that B happens given that A has already occurred.
Example 3.2.11.
A student in Modegashe primary school was instructed to roll a Die and Toss a Coin.
What was the probability of rolling a 4 on the die and getting head on the coin?
Solution.
-
Probability of rolling a 4 on a six-sided die is
\begin{gather*}
\frac{1}{6}
\end{gather*}
-
Probability of getting heads on the coin is
\begin{gather*}
= \frac{1}{2}
\end{gather*}
\begin{gather*}
\textbf{so, } \, \textbf{P(4 and H)} = \frac{1}{6} \times \frac{1}{2}
\end{gather*}
\begin{gather*}
= \frac{1}{12}
\end{gather*}
the probability of rolling a 4 and flipping heads is \(\frac{1}{12}\) or \(8.33\%\)
Example 3.2.12.
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)
Find the probability that the person catches the first bus.
Find the probability that the person misses the first bus but catches the second.
Find the probability that the person misses both buses.
Solution.
Probability of catching the first bus:
\begin{gather*}
\textbf{P(A) = 0.6}
\end{gather*}
Probability of missing the first bus
\begin{gather*}
\textbf{P}(\textbf{A}^{c}) \, = \, 1 - \textbf{P(A)} \, = \, 1 - 0.6 \, = \, 0.4
\end{gather*}
Probability of catching the second bus, given that the first bus was missed
\begin{gather*}
\textbf{P}(\textbf{B}|\textbf{A}^{c}) \, = \, 0.8
\end{gather*}
Probability of missing the second bus, given that the first bus was missed
\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*}
-
The probability of catching the first bus is directly given as
\begin{gather*}
\textbf{P(A)} \, = \, 0.6
\end{gather*}
The probability of catching the first bus is\(0.6\) or \(60\%\text{.}\)
-
Missing the first bus \(\textbf{A}^{c}\)
Catching the second bus \(\text{B}\)
Since these events are dependent, we use the multiplication rule
\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*}
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B} ) \, = \, 0.4 \times 0.8 \, = \, 0.32
\end{gather*}
The probability of missing the first bus but catching the second is \(0.32\) or \(32\%\text{.}\)
-
Missing the first bus \(\textbf{A}^{c}\)
Missing the second bus \(\textbf{B}^{c}\)
Again, using the multiplication rule
\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*}
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B}^{c} ) \, = \, 0.4 \times 0.2 \, = \, 0.08
\end{gather*}
The probability of missing both buses is \(0.08\) or \(8\%\text{.}\)
\begin{gather*}
\end{gather*}
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 and getting tails?
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?
4.
A farmer in Gakuonyo plants two seeds. Each seed has a 75% chance of germinating. What is the probability that both seeds germinate?
