Probability of Independent Events

Understanding Independent Events

In real life, we often encounter situations where the result of one event does not affect the result of another event. Imagine you throw a coin and a die simultaneously. Does the result of the coin affect the number that appears on the die? Of course not! These two events are independent.

Independent events are two or more events where the result of one event does not affect the probability of another event. If event A does not affect event B, and vice versa, then both events are independent.

As a simple illustration, think about today's weather and your math exam result tomorrow. Whether it's rainy or sunny today will not affect your exam score (unless you're late because of the rain, but that's another story!). These two events are statistically independent.

Characteristics of Independent Events

Not Mutually Affecting

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 events, 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.

Daily Life Examples

Some examples of independent events that are easy to understand:

Throwing two different coins simultaneously
Drawing cards from two separate decks
Exam results of two different students (without cheating!)
Weather conditions in two distant cities

Identifying Independent Events

To identify whether two events are independent, ask: "Does knowing the result of the first event provide information about the result of the second event?" If the answer is no, then both events are independent.

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 This Formula Works

Unlike events that affect each other, in independent events the conditional probability equals the regular probability. Since $P(A|B) = P(A)$ , then:

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

This is why we can directly multiply individual probabilities to get the joint probability.

Application in Calculations

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}

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

P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}

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_{die} = \{1, 2, 3, 4, 5, 6\}

S_{coin} = \{H, T\}

Event A: getting an even number on the die = $\{2, 4, 6\}$
Event B: getting tails on the coin = $\{T\}$

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}

P(B) = \frac{1}{2}

P(A \cap B) = P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Real-World Application Example

In a city, the probability that a fire truck is needed on a particular day is $0.98$ , while the probability that an ambulance is needed is $0.92$ . What is the probability that both vehicles are needed on the same day?

These probability numbers represent a scenario of a large city with high emergency activity levels.

Solution:

Event A: fire truck is needed
Event B: ambulance is needed

Both events can be considered independent because the need for a fire truck does not affect the need for an ambulance.

Detailed calculation:

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

P(A \cap B) = 0.98 \times 0.92 = 0.9016

So the probability that both vehicles are needed on the same day is $0.9016$ or $90.16\%$ .

Problem-Solving Strategies

Systematic Steps

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

Identify the events involved in the problem
Ensure events are independent by verifying there is no mutual influence
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

Practical Tips

Some strategies to facilitate understanding:

Visualize events with tree diagrams if necessary
Ensure independence by asking whether one event affects another
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 Events:

Throwing multiple coins simultaneously
Drawing cards with replacement
Exam results of different students
Weather conditions in separate locations

Events that are NOT Independent:

Drawing cards without replacement
A person's height and weight
Test scores of the same student in different subjects
Air temperature and humidity

Exercises

Two coins are thrown simultaneously. Find the probability of getting heads on the first coin and tails on the second coin.
A die and a card are drawn from a 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 coins are thrown simultaneously. Find the probability of getting exactly two tails.

Answer Key

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 = $\{2, 3, 5\}$ , 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

Sample space for 3 coins (H = Heads, T = Tails): $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$ , total = $8$

Event of exactly two tails: $\{HTT, THT, TTH\}$ , there are $3$ events

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

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 $\frac{3}{8}$ .

Command Palette