Compound events are combinations of two or more events within a sample space that are related to one another or occur simultaneously. In probability theory, compound events are often analyzed to determine the likelihood of two or more events occurring together. In general, compound events can be classified into several types: mutually exclusive events, non-mutually exclusive events, independent events, and dependent events.
Mutually exclusive events are two or more events that cannot occur at the same time. In other words, if one event occurs, the other event cannot occur. For example, in a single roll of a die, the event of obtaining an even number (2, 4, 6) and the event of obtaining an odd number (1, 3, 5) are mutually exclusive because a number cannot be both even and odd at the same time.
On the other hand, non-mutually exclusive events are events that can occur simultaneously. For instance, when drawing one card from a standard deck of cards, the event “drawing a red card” and the event “drawing a King” are not mutually exclusive because there are red Kings (the King of Hearts and the King of Diamonds), meaning both events can happen together.
Independent events are events that do not affect one another. An example is flipping a coin and rolling a die at the same time. The outcome of the coin toss (heads or tails) does not influence the result of the die roll (numbers 1 through 6), and vice versa.
Finally, dependent events are events that influence one another. For example, drawing two balls consecutively from a box without replacement. Once the first ball is drawn and not returned, the number of balls in the box decreases, causing the probability of the second event to change.
By understanding these types of compound events, we can calculate the probability of combined events more accurately according to the relationship between the events involved. To strengthen understanding of compound events, several practice problems and solutions related to this topic are presented below. The problems are designed at a high school level of difficulty, making them highly suitable for students who already possess a solid foundation in basic probability.
Wilson’s Law
Problem Number 1
From a bag containing 3 red balls, 2 yellow balls, and 5 green balls, two balls are randomly drawn one by one without replacement. Determine the probability that both balls drawn are green.
Notice that the first and second balls desired are both green. This means that the case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing a green ball on the first and second draws. Thus,
$$P(A) = \dfrac{\text{number of green balls}}{\text{total number of balls}} = \dfrac{5}{3+2+5} = \dfrac{5}{10} = \dfrac12$$and
$$P(B) =\dfrac{\text{number of green balls}}{\text{total number of balls}} = \dfrac{5-1}{3+2+5-1} = \dfrac49$$ because the number of green balls decreases by $1,$ as does the total number of balls. Furthermore,
$$\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac12 \times \dfrac49 = \dfrac29. \end{aligned}$$Therefore, the probability that both balls drawn are green is $\boxed{\dfrac29.}$
Problem Number 2
In a box, there are 10 cards numbered from 1 to 10. If one card is drawn, what is the probability of drawing a card with an odd number or a prime number?
Clearly, this is a compound event connected by the word “or”.
Suppose $A$ and $B$ respectively represent the events of drawing a card with an odd number and drawing a card with a prime number. $A = \{1, 3, 5, 7, 9\}$ and $B=\{2, 3, 5, 7\}$ are two non-mutually exclusive events because they can occur simultaneously (there are numbers that are both odd and prime), namely $A \cap B = \{3, 5, 7\}.$ Notice that $\text{n}(A) = 5,$ $\text{n}(B) = 4,$ $\text{n}(A \cap B) = 3,$ and the number of elements in the sample space is $\text{n}(S) = 10.$ Thus,
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{5}{10}$$and
$$P(B) =\dfrac{\text{n}(B)}{\text{n}(S)} = \dfrac{4}{10}$$as well as
$$P(A \cap B) = \dfrac{\text{n}(A \cap B)}{\text{n}(S)} = \dfrac{3}{10}$$so that
$$\begin{aligned} P(A \cup B) & = P(A)+P(B)-P(A \cap B) \\ & = \dfrac{5}{10} + \dfrac{4}{10}-\dfrac{3}{10} \\ & = \dfrac{6}{10} = \dfrac35. \end{aligned}$$Therefore, the probability of drawing a card with an odd number or a prime number is $\boxed{\dfrac35}.$
Problem Number 3
Inside a box, there are 4 red marbles and 6 blue marbles. Two marbles are drawn one by one without replacement. Calculate the probability of drawing a red marble on the first draw and a blue marble on the second draw.
Clearly, this case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing a red marble on the first draw and drawing a blue marble on the second draw. Thus,
$$P(A) = \dfrac{\text{number of red marbles}}{\text{total number of marbles}} = \dfrac{4}{4+6} = \dfrac{2}{5}$$and
$$P(B) =\dfrac{\text{number of blue marbles}}{\text{total number of marbles}} = \dfrac{6}{4+6-1} = \dfrac23$$ because $1$ red marble has been drawn. Furthermore,
$$\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac25 \times \dfrac23 = \dfrac{4}{15}. \end{aligned}$$Therefore, the probability of drawing a red marble on the first draw and a blue marble on the second draw is $\boxed{\dfrac{4}{15}}.$
Problem Number 4
In a box, there are 8 marbles consisting of 3 red, 2 blue, and 3 white marbles. Two marbles are drawn one by one without replacement. Calculate the probability that no blue marble is drawn.
Notice that the first and second marbles desired are marbles that are not blue. This means that the case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing a non-blue marble (that is, red or white) on the first and second draws. Thus,
$$P(A) = \dfrac{\text{number of red+white marbles}}{\text{total number of marbles}} = \dfrac{3+3}{3+2+3} = \dfrac{3}{4}$$and
$$P(B) =\dfrac{\text{number of red+white marbles}}{\text{total number of marbles}} = \dfrac{3+3-1}{3+2+3-1} = \dfrac{5}{7}$$ because $1$ non-blue marble has been drawn. Furthermore,
$$\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac34 \times \dfrac57 = \dfrac{15}{28}. \end{aligned}$$Therefore, the probability that no blue marble is drawn is $\boxed{\dfrac{15}{28}}.$
Problem Number 5
A box contains 10 cards numbered from 1 to 10. Two cards are drawn one by one without replacement. Determine the probability that both cards drawn are even-numbered.
Notice that the first and second cards desired are even-numbered. This means that the case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing an even-numbered card on the first and second draws. From $1$ to $10,$ there are a total of $5$ even numbers, namely $2, 4, 6, 8,$ and $10.$ Thus,
$$P(A) = \dfrac{\text{number of even-numbered cards}}{\text{total number of cards}} = \dfrac{5}{10} = \dfrac{1}{2}$$and
$$P(B) =\dfrac{\text{number of even-numbered cards}}{\text{total number of cards}} = \dfrac{5-1}{10-1} = \dfrac49$$ because $1$ even-numbered card has been drawn. Furthermore,
$$\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac12 \times \dfrac49 = \dfrac{2}{9}. \end{aligned}$$Therefore, the probability that both cards drawn are even-numbered is $\boxed{\dfrac29}.$
Problem Number 6
Inside a bag, there are 2 small dice colored red and blue. A die is rolled one by one. Calculate the probability that the first die shows an even number and the second die shows an odd number.
Clearly, this case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events that the first die shows an even number and the second die shows an odd number. There are $3$ even faces, namely $2, 4,$ and $6$, and likewise there are $3$ odd faces, namely $1, 3,$ and $5.$ Thus,
$$P(A) = \dfrac{\text{number of even die faces}}{\text{number of distinct die faces}} = \dfrac{3}{6} = \dfrac12$$and
$$P(B) =\dfrac{\text{number of odd die faces}}{\text{number of distinct die faces}} = \dfrac{3}{6} = \dfrac{1}{2}$$so that we obtain
\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac12 \times \dfrac12 = \dfrac{1}{4}. \end{aligned}$$Therefore, the probability that the first die shows an even number and the second die shows an odd number is$\boxed{\dfrac{1}{4}}.$
Problem Number 7
From a box containing 5 strawberry candies, 4 melon candies, and 3 grape candies, two candies are drawn one by one without replacement. Calculate the probability that the first candy is strawberry and the second candy is grape.
Clearly, this case is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing a strawberry candy first and a grape candy second. Thus,
$$P(A) = \dfrac{\text{number of strawberry candies}}{\text{total number of candies}} = \dfrac{5}{5+4+3} = \dfrac{5}{12}$$and
$$P(B) =\dfrac{\text{number of grape candies}}{\text{total number of candies}} = \dfrac{3}{5+4+3-1} = \dfrac{3}{11}$$because $1$ strawberry candy has been drawn. Furthermore,
\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac{5}{\cancelto{4}{12}} \times \dfrac{\cancel{3}}{11} = \dfrac{15}{44}. \end{aligned}$$Therefore, the probability that the first candy is strawberry and the second candy is grape is $\boxed{\dfrac{15}{44}}.$
Problem Number 8
A box contains 4 blue marbles and 6 white marbles. Two marbles are drawn one by one without replacement. Calculate the probability that the two marbles drawn have different colors.
Notice that the desired first and second marbles are marbles of different colors. This means that the case is a compound event connected by the word “and”.
There are two cases that need to be considered in this situation.
Case 1: The first marble is blue and the second marble is red
Suppose $A$ and $B$ respectively represent the events of drawing a blue marble first and a red marble second. Thus,
$$P(A) = \dfrac{\text{number of blue marbles}}{\text{total number of marbles}} = \dfrac{4}{4+6} = \dfrac{2}{5}$$and
$$P(B) =\dfrac{\text{number of red marbles}}{\text{total number of marbles}} = \dfrac{6}{4+6-1} = \dfrac{2}{3}$$because $1$ blue marble has been drawn. Furthermore,
\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac{2}{5} \times \dfrac{2}{3} = \dfrac{4}{15}. \end{aligned}$$Case 2: The first marble is red and the second marble is blue
Suppose $C$ and $D$ respectively represent the events of drawing a red marble first and a blue marble second. Thus,
$$P(C) = \dfrac{\text{number of red marbles}}{\text{total number of marbles}} = \dfrac{6}{4+6} = \dfrac{3}{5}$$and
$$P(D) =\dfrac{\text{number of blue marbles}}{\text{total number of marbles}} = \dfrac{4}{4+6-1} = \dfrac{4}{9}$$because $1$ red marble has been drawn. Furthermore,
\begin{aligned} P(C \cap D)& = P(C) \times P(D) \\ & = \dfrac{\cancel{3}}{5} \times \dfrac{4}{\cancelto{3}{9}} = \dfrac{4}{15}. \end{aligned}$$Therefore, the probability that the two marbles drawn have different colors is $\boxed{\dfrac{4}{15} + \dfrac{4}{15} = \dfrac{8}{15}}.$
Problem Number 9
In a box, there are 4 red balls and 3 blue balls. One ball is drawn at random, then without replacement, another ball is drawn. Determine the probability of drawing at least one blue ball.
There are three possible cases if the desired outcome is drawing at least one blue ball. First, the first ball is red and the second ball is blue. Second, the opposite case, namely the first ball is blue and the second ball is red. Finally, both balls drawn are blue. The only possibility that does not satisfy the condition of at least one blue ball is when the first ball is red and the second ball is also red. We will determine the probability of drawing a red ball first and another red ball second so that the calculation becomes easier later by using the complement property in probability.
Clearly, this is a compound event connected by the word “and”.
Suppose $A$ and $B$ respectively represent the events of drawing a red ball first and a red ball second. Thus,
$$P(A) = \dfrac{\text{number of red balls}}{\text{total number of balls}} = \dfrac{4}{7}$$and
$$P(B) =\dfrac{\text{number of red balls}}{\text{total number of balls}} = \dfrac{4-1}{7-1} = \dfrac{3}{6} = \dfrac12$$because $1$ red ball has been drawn. Furthermore,
$$\begin{aligned} P(A \cap B)& = P(A) \times P(B) \\ & = \dfrac{\cancelto{2}{4}}{7} \times \dfrac{1}{\cancel{2}} = \dfrac{2}{7}. \end{aligned}$$We obtain that the probability of drawing a red ball first and another red ball second is $\dfrac27.$ To find the probability of drawing at least one blue ball, we simply subtract $\dfrac27$ from $1$ (the total probability). Mathematically, this is written as
$$P(\overline{A \cap B}) = 1-P(A \cap B) = 1-\dfrac27= \dfrac57.$$ Therefore, the probability of drawing at least one blue ball is $\boxed{\dfrac57}.$
Problem Number 10
Determine the number of subsets of $\{1, 2, 3, 4, 5, 6, 7\}$ that contain the set $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6\}.$
Suppose $S = \{1, 2, 4, 4, 5, 6, 7\}.$ Also suppose that $A$ represents the subsets of $S$ containing $\{1, 2, 3, 4, 5\},$ while $B$ represents the subsets of $S$ containing $\{4, 5, 6\}.$ In this case, we want to determine
$$|A \cup B| = |A|+|B|-|A \cap B|.$$Step 1: Determine the value of $|A|.$
Notice that $A = \{1, 2, 3, 4, 5\} \cup X$ with $X \subseteq \{6, 7\}.$ Each element of $X$ has two possibilities, either included or not included. Since $X$ consists of two elements, the number of possibilities is $|A| = |X| = 2^2 = 4,$ namely $X = \{ \},$ $X = \{6\},$ $X = \{7\},$ and $X = \{6, 7\}.$
Step 2: Determine the value of $|B|.$
Notice that $B = \{4, 5, 6\} \cup Y$ with $Y \subseteq \{1, 2, 3, 7\}.$ Each element of $Y$ has two possibilities, either included or not included. Since $Y$ consists of four elements, the number of possibilities is $|B| = |Y| = 2^4 = 16.$
Step 3: Determine the value of $|A \cup B|.$
Notice that $A \cup B = \{1, 2, 3, 4, 5, 6\} \cup Z$ with $Z \subseteq \{7\}.$ Each element of $Z$ has two possibilities, either included or not included. Since $Z$ consists of one element, the number of possibilities is $|A \cup B| = |Z| = 2^1 = 2.$
Thus, we obtain
$$\begin{aligned} |A \cup B| & = |A|+|B|-|A \cap B| \\ & = 4 + 16 -2 = 18. \end{aligned}$$Therefore, the number of subsets of $\{1, 2, 3, 4, 5, 6, 7\}$ that contain the set $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6\}$ is $\boxed{18}.$