Solution to Example 2
Two methods to answer the question in example 2 are presented to show the advantage in using the product rule given above.
Method 1: Using the sample space
The sample space S of the experiment of tossing a coin twice is given by the tree diagram shown below
The first toss gives two possible outcomes : T or H ( in blue)
The second toss gives two possible outcomes : T or H (in red)
From the tree diagram, we can deduce the sample space \( S \) set as follows
\( S = \{(H, H), (H, T), (T, H), (T,T) \} \)
with \( n(S) = 4 \) where \( n(S) \) is the number of elements in the set \( S \)
The event \( E \) : " tossing a coin twice and getting two tails " as a set is given by
\( E = \{(T,T) \} \)
with \( n(E) = 1 \) where \( n(E) \) is the number of elements in the set \( E \)
Use the classical probability formula to find \( P(E) \) as:
\( P(E) = \dfrac{n(E)}{n(S)} = \dfrac{1}{4}\)
Method 2: Use the product rule of two independent event
Event \( E \) " tossing a coin twice and getting a tail in each toss " may be considered as two events
Event \( A \) " toss a coin once and get a tail " and event \( B \) "toss the coin a second time and get a tail "
with the probabilities of each event \( A \) and \(B \) given by
\( P(A) = \dfrac{1}{2} \) and \( P(B) = \dfrac{1}{2} \)
Event E occurring may now be considered as events A and B occurring. Events A and B are independent and therefore the product rule may be used as follows
\( P(E) = P( A \; and \; B) = P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4} \)
NOTE If you toss a coin a large number of times, the sample space will have a large number of elements and therefore method 2 is much more practical to use than method 1 where you have a large number of outcomes.
We now present more examples and questions on how the product rule of independent events is used to solve probability questions.
Example 3
A coin is tossed and a die is rolled. What is the probability of getting a head and a \( 4 \)?
Solution to Example 3
We have two independent events to consider:
Event A " toss a coin a get a head " and event B " roll a die and get a \( 4 \) "
When a coin is tossed, the probability of getting a head is
\( P(A) = \dfrac{1}{2} \)
When die is rolled, the probability of getting a \( 4 \) is
\( P(B) = \dfrac{1}{6} \)
\( P ( \) " getting a head and a \( 4 \) " \( ) = P( A \; and \; B) = P( A \cap B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{6} = \dfrac{1}{12} \)
Example 4
A jar has 3 blue balls, 2 white balls and 5 red balls. A ball is selected at random and the color noted then replaced back inside the jar. A second ball is selected its color noted and replaced back inside the jar. A third ball is selected and its color noted.
What is the probability of
a) selecting 3 red balls
b) selecting a blue ball, then a white ball and then a blue ball
c) selecting a red, then a white ball and then a blue ball
Solution to Example 4
a)
Let event A "select a red ball the first time",
event B "select a red ball the second time"
and event C "select a red ball the third time"
All three events A, B and C are independent because the ball selected is replaced back into the jar.
Total number of balls is 10 and there are 5 red balls.
Let us now calculate the probability of selecting a red ball.
There are 5 red balls out of a total of 10, hence
\( P(A) \) = \( P(B) \) = \( P(C) \) =\( P(red) = \dfrac{5}{10} \\ = \dfrac{1}{2} \)
We use an extended formula to three independent events
\( P( \; A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\ = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{8} \)
b)
Let event A "select a blue ball the first time",
event B "select a white ball the second time"
and event C "select a blue ball the third time"
\( P(A) = P(blue) = \dfrac{3}{10} \quad , \quad P(B) = P(white) = \dfrac{2}{10} = \dfrac{1}{5} \quad , \quad P(C) = P(blue) = \dfrac{3}{10}\)
\( P( A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\ = \dfrac{3}{10} \cdot \dfrac{1}{5} \cdot \dfrac{3}{10} \dfrac{3}{10} \\\\ = \dfrac{9}{5000} \)
c)
Let event A "select a red ball the first time",
event B "select a white ball the second time"
and event C "select a blue ball the third time"
\( P(A) = P(red) = 1/2 \)
\( P(B) = P(white) = 1/5 \)
\( P(C) = P(blue) = 3/10 \)
\( P( A \; and \; B \; and \; C) = P(A) \cdot P(B) \cdot P(C) \\\\ = \dfrac{1}{2} \cdot \dfrac{1}{5} \cdot \dfrac{3}{10} \\\\ = \dfrac{3}{100} \)
Example 5
A card is drawn from a deck of 52 cards and then replaced and a second card is drawn. Find the probability of getting a "2" and then a "King".
Solution to Example 5
We have two independent events to consider:
Event A " draw a card and get a 2" and event B " draw a card and get a King "
Because the card is replaced, the two events A and B are independent.
Let us first find \( P(A) \) and \( P(B) \).
\( P(A) = 4/52 = 1/13 \)
\( P(B) = 4/52 = 1/13 \)
\( P (A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{13} = \dfrac{1}{169} \)
