Multiplication Rule for Probabilities of Independent Events

# Multiplication Rule for Probabilities of Independent Events

Examples on using the multiplication rule to find the probability of two or more independent events occurring are presented along with detailed solutions.

## Independent Events

In probabilities, two events are independent if the occurence of one does not affect the probability of occurence of the other.
Example 1
The follwing events A and B independent.

1. A = "roll a die and get a $1$" , B = "flip a coin and get a tail".
2. A = "draw a card from a deck and get a King", replace it back into the deck, B = "draw another card and get a Queen"
3. A = "roll a die and get a $4$" , B = "roll the same die (or another one) and get a "6"
4. A = "flip a coin and get a head" , B = roll the same coin (or another one) and get a tail"
A jar has 3 blue balls, 2 white balls and 5 red balls
5. A = Pick a ball at random from the jar and get a red ball, replace it back into the jar, B = Pick a ball at random from the jar and get a white ball

The events C and D are NOT independent.
1. C = "draw a card from a deck and get a King", D = "draw a second card from the same deck and get a Queen".
A jar contains 3 blue balls, 2 white balls and 3 red balls
2. C = "Pick a ball at random from the jar and get a red ball", D = "Pick a second ball at random from the same jar and get a white ball".
The probability of two independent events A and B both occurring is given by the product of the probability of each event occurring. $P(A \; \text{and} \; B) = P(A)\cdot P(B)$ or using the set notation $P(A \cap B) = P(A)\cdot P(B)$

## Examples with Detailed Solutions

Example 2 A coin is tossed twice. What is the probability of getting a tail in the first toss and a tail in the second toss?

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 three 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 $P(B)$.

$P(A) = 4/52 = 1/13$
$P(B) = 4/52 = 1/13$
$P (A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{13} = \dfrac{1}{169}$

Example 6
A poll found that 25% of the people in a certain country have heart problems. If three people are selected at random, find the probability that all three have heart problems.

Solution to Example 6
Event A " the first person have heart problems" , event B " the second person have heart problems " and C " the third person have heart problems "
$P(A) = 0.25$ , $P(B) = 0.25$ and $P(C) = 0.25$.
These are independent events, hence
$P (A and B and C ) = P(A) \cdot P(B) \cdot P(C) = 0.25 \cdot 0.25 \cdot 0.25 = 0.015625$

### More Questions with Solutions

1) A die is rolled twice. Find the probability of getting an even number in the first throw and a number greater than 4 in the second throw.
2) 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 "King" and then a "Queen of hearts".
3) In a country, 45% of the people smoke. If 4 people are selected at random, what is the population that they are all smokers?

### Solutions to above exercises

1)
Let event A: get an even number , event B: get a number greater than 4
A = $\{2,4,6\}$ , B = $\{5,6\}$
$P(A) = 3/6 = 1/2$ , $P(B) = 2/6 = 1/3$
Events A and B are independent; hence
$P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$
2)
$P(King) = P(A) = 4/52 = 1/13$ , $P(Queen \; of \; hearts) = P(B) = 1/52$
Events A and B are independent; hence
$P(A \; and \; B) = P(A) \cdot P(B) = \dfrac{1}{13} \cdot \dfrac{1}{52} = \dfrac{1}{676}$
3)
$P(smoker) = 0.45$
All 4 events are independent; hence
$P(\text{all 4 smokers}) = P(smoker \; and \; smoker \; and \; smoker \; and \; smoker) \\\\ = P(smoker) \cdot P(smoker) \cdot P(smoker) \cdot P(smoker) \\\\ = (0.45)^4 = 0.04100625$