Probability of Independent Events

Learning One Outcome Leaves the Other Probability Unchanged

Some probability models contain events where learning one result does not change the probability of another. Toss a fair coin and roll a fair die. Knowing the coin result gives no information about the die result, so the two events are independent.

Events $A$ and $B$ are independent when conditioning on either event leaves the probability of the other unchanged. The probability model determines this property. Events that happen at the same time are not automatically independent.

Events can share a hidden cause even when neither event directly controls the other. Two alarms, for example, can both react to the same incident. Treat events as independent only when the probability model states it or the conditional probabilities confirm it.

How Independence Shows Up in Probabilities

Independence leaves the second probability unchanged once the first outcome is known, and it also requires separate trials. The three checks below state that idea in different forms.

Unchanged Conditional Probability

The main characteristic of independent events is that the result of one event does not change the probability of another event. In mathematical notation, if $A$ and $B$ are independent and $P(B) > 0$ , then:

P(A|B) = P(A)

This means the probability of $A$ occurring when $B$ has already occurred is the same as the probability of $A$ occurring in general.

Separate Random Trials

Examples with a clear independent mechanism include:

tossing two separate fair coins
drawing one card from each of two independently shuffled decks
rolling two separate fair dice
generating two random digits with independent trials

Identifying Independent Events

To test independence, ask: “Does knowing the first result change the probability distribution of the second?” A suspected causal link alone is not enough. Use the stated model or compare the conditional probability with the ordinary probability.

Formula for Probability of Independent Events

Since independent events do not affect each other, calculating their joint probability becomes simple. The basic formula for the probability of independent events is:

P(A \cap B) = P(A) \times P(B)

This formula shows that the probability of both events $A$ and $B$ occurring together equals the multiplication of the probability of each event.

Why Multiplying the Probabilities Works

For independent events, the conditional probability equals the ordinary probability. Because $P(A|B) = P(A)$ , the multiplication rule becomes:

P(A \cap B) = P(A|B) \times P(B) = P(A) \times P(B)

Substituting $P(A|B)=P(A)$ into the multiplication rule gives $P(A\cap B)=P(A)P(B)$ .

Calculations with Independent Events

Each calculation below multiplies the probabilities of two independent events. The examples move from two dice to a die and a coin, then to two sensor self tests.

Rolling Two Dice

Two dice are rolled simultaneously, one red die and one white die. Find the probability of getting number $2$ on the red die and number $5$ on the white die.

Solution:

Event $A$ : getting number $2$ on the red die
Event $B$ : getting number $5$ on the white die

Both events are independent because the result of the red die does not affect the result of the white die.

P(A) = \frac{1}{6}

So the probability of getting number $2$ on the red die and number $5$ on the white die is $\frac{1}{36}$ .

Rolling a Die and Coin

A die and a coin are thrown simultaneously. Find the probability of getting an even number on the die and tails on the coin.

Solution:

S_1 = \{1, 2, 3, 4, 5, 6\}

Event $A$ : getting an even number on the die is $\{2, 4, 6\}$
Event $B$ : getting tails on the coin, represented by $1$ , is $\{1\}$

Both events are independent because the result of the die does not affect the result of the coin.

P(A) = \frac{3}{6} = \frac{1}{2}

Independent Sensor Self Tests

Two independently powered monitoring sensors run a daily self-test. Sensor $A$ succeeds with probability $0.98$ , while sensor $B$ succeeds with probability $0.92$ . Under the stated independence model, what is the probability that both tests succeed?

The independence assumption is part of this model. If both sensors shared a power supply or network, we would need to test that assumption again.

Solution:

Event $A$ : the first sensor passes its self-test
Event $B$ : the second sensor passes its self-test

The problem states that the two self-tests are independent, so one result does not change the probability of the other.

Detailed calculation:

P(A \cap B) = P(A) \times P(B)

So the probability that both sensors pass is $0.9016$ or $90.16\%$ .

How to Work Through an Independence Problem

Most mistakes in these problems come from multiplying before checking independence. The steps below put that check first and end by verifying the answer. If the check fails, the multiplication rule for independent events does not apply at all.

Check Independence Before Multiplying

To solve probability problems involving independent events, follow these steps:

Identify the events involved in the problem
Verify independence from the stated model or the conditional probabilities
Calculate individual probabilities of each event
Apply the formula $P(A \cap B) = P(A) \times P(B)$
Check the result to see if it makes sense in the problem context

Checks for the Final Answer

Use these checks:

Visualize events with tree diagrams if necessary
Verify independence from the model or conditional probabilities. Timing alone does not prove it
Check consistency of results using different approaches
Use context to validate whether the answer makes sense

Recognizing Independent Events

Pay attention to the following characteristics to identify independent events:

Independent under the stated model:

separate fair-coin tosses
card draws with replacement and reshuffling
rolls of separate fair dice
independent random-number trials

Situations where independence must not be assumed:

Drawing cards without replacement
A person's height and weight, unless a model or data justify independence
Test scores of the same student in different subjects, unless independence is established
Air temperature and humidity, unless independence is established for the stated model

Exercises

Two fair coins are tossed independently. Find the probability of getting heads on the first coin and tails on the second coin.
A fair die is rolled independently of drawing one card from a shuffled standard deck. Calculate the probability of getting a prime number on the die and a red card.
In a class, the probability of a student passing mathematics is $0.85$ and the probability of passing physics is $0.78$ . If both subjects are independent, what is the probability that the student passes both subjects?
Three fair coins are tossed independently. Find the probability of getting exactly two tails.

Try each exercise before reading its worked solution. State why the events are independent before multiplying their probabilities.

Worked Solutions

Solution:
- Event $A$ : heads on the first coin, $P(A) = \frac{1}{2}$
- Event $B$ : tails on the second coin, $P(B) = \frac{1}{2}$
Both events are independent because the result of the first coin does not affect the second coin.

$P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Solution:
- Event $A$ : prime number on the die is $\{2, 3, 5\}$ , so $P(A) = \frac{3}{6} = \frac{1}{2}$
- Event $B$ : red card (hearts and diamonds), so $P(B) = \frac{26}{52} = \frac{1}{2}$
Both events are independent.

$P(A \cap B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Solution:
- Event $A$ : passing mathematics, $P(A) = 0.85$
- Event $B$ : passing physics, $P(B) = 0.78$
Since both subjects are independent:

$P(A \cap B) = 0.85 \times 0.78 = 0.663$

So the probability of passing both subjects is $0.663$ or $66.3\%$ .
Solution:

Method 1: Direct Enumeration

Use $0$ for heads and $1$ for tails. The sample space for $3$ coins contains $8$ outcomes:

$\left\{\begin{gathered} 000, 001, 010, 011, \\ 100, 101, 110, 111 \end{gathered}\right\}$

Exactly two tails occurs in $\{011, 101, 110\}$ , so there are $3$ favorable outcomes.

$P(\text{exactly 2 tails}) = \frac{3}{8}$

Method 2: Using Binomial Distribution

For $n = 3$ tosses with success probability (tails) $p = \frac{1}{2}$ , the probability of exactly $k = 2$ successes is:

$P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}$

Substituting values:

$P(\text{exactly 2 tails}) = \binom{3}{2} \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^1$

Step-by-step explanation:
- $\binom{3}{2} = 3$ (ways to choose $2$ positions out of $3$ for tails)
- $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ (probability of $2$ tails)
- $\left(\frac{1}{2}\right)^1 = \frac{1}{2}$ (probability of $1$ head)
So, the calculation is:

$P(\text{exactly 2 tails}) = 3 \times \frac{1}{4} \times \frac{1}{2} = 3 \times \frac{1}{8} = \frac{3}{8}$

So the probability of getting exactly two tails is $\frac{3}{8}$ .

Learning One Outcome Leaves the Other Probability Unchanged

How Independence Shows Up in Probabilities

Independence leaves the second probability unchanged once the first outcome is known, and it also requires separate trials. The three checks below state that idea in different forms.

Unchanged Conditional Probability

P(A|B) = P(A)

This means the probability of $A$ occurring when $B$ has already occurred is the same as the probability of $A$ occurring in general.

Separate Random Trials

Examples with a clear independent mechanism include:

tossing two separate fair coins
drawing one card from each of two independently shuffled decks
rolling two separate fair dice
generating two random digits with independent trials

Identifying Independent Events

Formula for Probability of Independent Events

Since independent events do not affect each other, calculating their joint probability becomes simple. The basic formula for the probability of independent events is:

P(A \cap B) = P(A) \times P(B)

This formula shows that the probability of both events $A$ and $B$ occurring together equals the multiplication of the probability of each event.

Why Multiplying the Probabilities Works

For independent events, the conditional probability equals the ordinary probability. Because $P(A|B) = P(A)$ , the multiplication rule becomes:

P(A \cap B) = P(A|B) \times P(B) = P(A) \times P(B)

Substituting $P(A|B)=P(A)$ into the multiplication rule gives $P(A\cap B)=P(A)P(B)$ .

Calculations with Independent Events

Each calculation below multiplies the probabilities of two independent events. The examples move from two dice to a die and a coin, then to two sensor self tests.

Rolling Two Dice

Two dice are rolled simultaneously, one red die and one white die. Find the probability of getting number $2$ on the red die and number $5$ on the white die.

Solution:

Event $A$ : getting number $2$ on the red die
Event $B$ : getting number $5$ on the white die

Both events are independent because the result of the red die does not affect the result of the white die.

P(A) = \frac{1}{6}

So the probability of getting number $2$ on the red die and number $5$ on the white die is $\frac{1}{36}$ .

Rolling a Die and Coin

A die and a coin are thrown simultaneously. Find the probability of getting an even number on the die and tails on the coin.

Solution:

S_1 = \{1, 2, 3, 4, 5, 6\}

Event $A$ : getting an even number on the die is $\{2, 4, 6\}$
Event $B$ : getting tails on the coin, represented by $1$ , is $\{1\}$

Both events are independent because the result of the die does not affect the result of the coin.

P(A) = \frac{3}{6} = \frac{1}{2}

Independent Sensor Self Tests

The independence assumption is part of this model. If both sensors shared a power supply or network, we would need to test that assumption again.

Solution:

Event $A$ : the first sensor passes its self-test
Event $B$ : the second sensor passes its self-test

The problem states that the two self-tests are independent, so one result does not change the probability of the other.

Detailed calculation:

P(A \cap B) = P(A) \times P(B)

So the probability that both sensors pass is $0.9016$ or $90.16\%$ .

How to Work Through an Independence Problem

Check Independence Before Multiplying

To solve probability problems involving independent events, follow these steps:

Identify the events involved in the problem
Verify independence from the stated model or the conditional probabilities
Calculate individual probabilities of each event
Apply the formula $P(A \cap B) = P(A) \times P(B)$
Check the result to see if it makes sense in the problem context

Checks for the Final Answer

Use these checks:

Visualize events with tree diagrams if necessary
Verify independence from the model or conditional probabilities. Timing alone does not prove it
Check consistency of results using different approaches
Use context to validate whether the answer makes sense

Recognizing Independent Events

Pay attention to the following characteristics to identify independent events:

Independent under the stated model:

separate fair-coin tosses
card draws with replacement and reshuffling
rolls of separate fair dice
independent random-number trials

Situations where independence must not be assumed:

Drawing cards without replacement
A person's height and weight, unless a model or data justify independence
Test scores of the same student in different subjects, unless independence is established
Air temperature and humidity, unless independence is established for the stated model

Exercises

Two fair coins are tossed independently. Find the probability of getting heads on the first coin and tails on the second coin.
A fair die is rolled independently of drawing one card from a shuffled standard deck. Calculate the probability of getting a prime number on the die and a red card.
In a class, the probability of a student passing mathematics is $0.85$ and the probability of passing physics is $0.78$ . If both subjects are independent, what is the probability that the student passes both subjects?
Three fair coins are tossed independently. Find the probability of getting exactly two tails.

Try each exercise before reading its worked solution. State why the events are independent before multiplying their probabilities.

Worked Solutions

Solution:
- Event $A$ : heads on the first coin, $P(A) = \frac{1}{2}$
- Event $B$ : tails on the second coin, $P(B) = \frac{1}{2}$
Both events are independent because the result of the first coin does not affect the second coin.

$P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Solution:
- Event $A$ : prime number on the die is $\{2, 3, 5\}$ , so $P(A) = \frac{3}{6} = \frac{1}{2}$
- Event $B$ : red card (hearts and diamonds), so $P(B) = \frac{26}{52} = \frac{1}{2}$
Both events are independent.

$P(A \cap B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Solution:
- Event $A$ : passing mathematics, $P(A) = 0.85$
- Event $B$ : passing physics, $P(B) = 0.78$
Since both subjects are independent:

$P(A \cap B) = 0.85 \times 0.78 = 0.663$

So the probability of passing both subjects is $0.663$ or $66.3\%$ .
Solution:

Method 1: Direct Enumeration

Use $0$ for heads and $1$ for tails. The sample space for $3$ coins contains $8$ outcomes:

$\left\{\begin{gathered} 000, 001, 010, 011, \\ 100, 101, 110, 111 \end{gathered}\right\}$

Exactly two tails occurs in $\{011, 101, 110\}$ , so there are $3$ favorable outcomes.

$P(\text{exactly 2 tails}) = \frac{3}{8}$

Method 2: Using Binomial Distribution

For $n = 3$ tosses with success probability (tails) $p = \frac{1}{2}$ , the probability of exactly $k = 2$ successes is:

$P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}$

Substituting values:

$P(\text{exactly 2 tails}) = \binom{3}{2} \times \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{2}\right)^1$

Step-by-step explanation:
- $\binom{3}{2} = 3$ (ways to choose $2$ positions out of $3$ for tails)
- $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ (probability of $2$ tails)
- $\left(\frac{1}{2}\right)^1 = \frac{1}{2}$ (probability of $1$ head)
So, the calculation is:

$P(\text{exactly 2 tails}) = 3 \times \frac{1}{4} \times \frac{1}{2} = 3 \times \frac{1}{8} = \frac{3}{8}$

So the probability of getting exactly two tails is $\frac{3}{8}$ .