The Multiplication Rule – Bernoulli Trials Revision Notes for Leaving Cert Mathematics

The Multiplication Rule – Bernoulli Trials

Introduction to the multiplication rule

When dealing with probability problems involving two or more events, we often need to find the probability that all events occur together. The multiplication rule provides us with a systematic way to calculate these combined probabilities.

Consider the simple example of tossing a coin and throwing a dice simultaneously. We can create a sample space showing all possible outcomes:

infoNote

Understanding Sample Spaces

The sample space method involves listing all possible outcomes to calculate probabilities. While effective, it can become cumbersome with larger numbers of events. The multiplication rule offers a more efficient alternative.

Coin	Dice Results
	1	2	3	4	5	6
H	H1	H2	H3	H4	H5	H6
T	T1	T2	T3	T4	T5	T6

From this sample space, we can see that there are 12 equally likely outcomes. The probability of getting both a head and a 6 is:

$P(H, 6) = \frac{1}{12}$

However, we can reach the same result more efficiently using the multiplication rule.

The multiplication rule formula

chatImportant

The Multiplication Rule (AND Rule)

When two events are independent (meaning the outcome of one event doesn't affect the outcome of the other), we can use the following formula:

$\boxed{P(A \text{ and } B) = P(A) \times P(B)}$

This rule is also commonly referred to as the AND rule.

Using our coin and dice example:

The probability of getting a head when tossing a coin is $\frac{1}{2}$
The probability of getting a 6 on the dice is $\frac{1}{6}$
Since these events are independent, $P(\text{head and 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

This matches our result from the sample space method, demonstrating the effectiveness of the multiplication rule.

Worked examples using the multiplication rule

lightbulbExample

Worked Example 1: Spinner and Dice

Amanda throws an ordinary dice and spins a spinner where each colour is equally likely. We need to find the probability that she gets a red and an even number.

Solution:

$P(\text{red}) = \frac{1}{3}$ (assuming 3 equal sections on spinner)
$P(\text{even number}) = \frac{3}{6} = \frac{1}{2}$ (even numbers are 2, 4, 6)
$P(\text{red and even number}) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$

lightbulbExample

Worked Example 2: Birthday Probability

Mary and John have their birthdays in the same week. Find the probability that: (i) Mary's birthday falls on Monday (ii) Both have their birthdays on Monday
(iii) Both have their birthdays on either Saturday or Sunday

Solution: (i) $P(\text{Mary's birthday on Monday}) = \frac{1}{7}$

(ii) $P(\text{both birthdays on Monday}) = P(\text{Mary on Monday}) \times P(\text{John on Monday}) = \frac{1}{7} \times \frac{1}{7} = \frac{1}{49}$

(iii) $P(\text{both birthdays on Sat or Sun}) = \frac{2}{7} \times \frac{2}{7} = \frac{4}{49}$

Introduction to Bernoulli trials

Consider the experiment of throwing a dice and requiring a 6 to start a game. We can think of getting a 6 as a success and any other number as a failure.

If each throw of the dice is regarded as a trial, then we notice three important characteristics:

For each trial, there are exactly two possible outcomes: 'success' and 'failure'
The probability of success (getting a 6) remains the same for each trial: $\frac{1}{6}$
Each trial is independent of the outcomes of other trials

When an experiment consists of repeated trials that satisfy these conditions, such trials are known as Bernoulli trials, named after James Bernoulli.

infoNote

Historical Context: James Bernoulli

James Bernoulli (1654-1705) was a Swiss mathematician who did pioneering work in probability and calculus, laying the foundation for much of modern probability theory. His contributions to the field of probability have made him one of the most influential mathematicians in this area.

Characteristics of Bernoulli trials

chatImportant

Three Essential Conditions for Bernoulli Trials

For trials to be classified as Bernoulli trials, they must satisfy three conditions:

Two outcomes only: Each trial has exactly two possible outcomes, typically called 'success' and 'failure'
Constant probability: The probability of success remains the same for each trial
Independence: Each trial is independent of the outcomes of other trials

Worked examples of Bernoulli trials

lightbulbExample

Worked Example 3: Coin Tossing Until First Head

A fair coin is tossed until a head occurs. Find the probability that the first head occurs on the third toss.

Solution: If the first head occurs on the 3rd toss, then the first two tosses must show tails.

The sequence is: TTH

$P(\text{T}) = \frac{1}{2}$ and $P(\text{H}) = \frac{1}{2}$
$P(\text{TTH}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

Therefore, the probability that a head first occurs on the 3rd toss is $\frac{1}{8}$ .

lightbulbExample

Worked Example 4: Dice Rolling Until Success

A fair dice is rolled repeatedly. Find the probability that a 5 or a 6 first appears on the third throw.

Solution: Let success (S) = getting a 5 or 6, and failure (F) = getting 1, 2, 3, or 4.

$P(\text{success}) = \frac{2}{6} = \frac{1}{3}$ and $P(\text{failure}) = \frac{4}{6} = \frac{2}{3}$

If the first 'success' is on the 3rd throw, the sequence is: FFS

$P(\text{FFS}) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$

Therefore, the probability that 5 or 6 first appears on the 3rd throw is $\frac{4}{27}$ .

lightbulbExample

Worked Example 5: Multiple Choice Test

A candidate takes a 3-question multiple choice test with four choices in each question. If she guesses on each question, what is the probability that: (i) She gets all three answers correct (ii) She gets the first two answers wrong but the third correct?

Solution: There are 4 choices per question, so:

$P(\text{correct answer}) = \frac{1}{4}$ and $P(\text{wrong answer}) = \frac{3}{4}$

(i) $P(\text{all three correct}) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}$

(ii) $P(\text{first 2 wrong and 3rd correct}) = P(\text{wrong}) \times P(\text{wrong}) \times P(\text{correct})$ $= \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{9}{64}$

Important note about Bernoulli trials

chatImportant

Common Mistake to Avoid

When we have exactly three Bernoulli trials and want to find the probability of getting exactly one success, we need to consider all possible positions where that success can occur:

$P(\text{exactly one success in three trials}) = P(\text{FFS}) + P(\text{FSF}) + P(\text{SFF})$

This accounts for the success occurring on the first, second, or third trial respectively. Don't forget to include all possible arrangements!

bookmarkSummary

Key Points to Remember:

The multiplication rule states that for independent events: $P(A \text{ and } B) = P(A) \times P(B)$
The multiplication rule is also called the AND rule and provides an alternative to using sample spaces
Bernoulli trials are repeated experiments where each trial has exactly two outcomes (success/failure), constant probability of success, and independence between trials
When calculating Bernoulli trial probabilities, multiply the individual probabilities for each trial in the sequence
For problems involving "first success on trial n", remember that all previous trials must be failures

The Multiplication Rule – Bernoulli Trials (Leaving Cert Mathematics): Revision Notes