Understanding Probability and Events

The problem asks us to determine the relationship between three events related to the digits of numbers printed on soccer balls. We need to analyze if these events are independent, dependent, pairwise independent, or mutually exclusive.

Defining the Sample Space

The box contains four soccer balls with numbers: 112, 121, 211, and 222. When a footballer chooses one ball at random, the set of all possible outcomes, known as the sample space ($\Omega$), is:

• $\Omega = \{112, 121, 211, 222\}$

The total number of outcomes is $|\Omega| = 4$.

Defining the Events

The problem defines three events:

• Event A\(_1\): The first digit of the printed number is 1.
• Event A\(_2\): The second digit of the printed number is 1.
• Event A\(_3\): The third digit of the printed number is 1.

Let's list the outcomes corresponding to each event:

• For A\(_1\), the numbers are 112 and 121 (first digit is 1). So, A\(_1\) = {112, 121}.
• For A\(_2\), the numbers are 112 and 211 (second digit is 1). So, A\(_2\) = {112, 211}.
• For A\(_3\), the numbers are 121 and 211 (third digit is 1). So, A\(_3\) = {121, 211}.

Calculating Probabilities of Individual Events

Since each ball is chosen at random, each outcome is equally likely. The probability of an event is the number of outcomes in the event divided by the total number of outcomes in the sample space.

• P(A\(_1\)) = $\frac{|A_1|}{|\Omega|} = \frac{2}{4} = \frac{1}{2}$
• P(A\(_2\)) = $\frac{|A_2|}{|\Omega|} = \frac{2}{4} = \frac{1}{2}$
• P(A\(_3\)) = $\frac{|A_3|}{|\Omega|} = \frac{2}{4} = \frac{1}{2}$

Calculating Probabilities of Intersections

Now, let's find the intersections of these events and their probabilities.

• A\(_1\) ∩ A\(_2\): First digit is 1 AND second digit is 1. The number is 112. A\(_1\) ∩ A\(_2\) = {112}.
• A\(_1\) ∩ A\(_3\): First digit is 1 AND third digit is 1. The number is 121. A\(_1\) ∩ A\(_3\) = {121}.
• A\(_2\) ∩ A\(_3\): Second digit is 1 AND third digit is 1. The number is 211. A\(_2\) ∩ A\(_3\) = {211}.
• A\(_1\) ∩ A\(_2\) ∩ A\(_3\): First digit is 1 AND second digit is 1 AND third digit is 1. No number in the sample space has all three digits as 1. A\(_1\) ∩ A\(_2\) ∩ A\(_3\) = ∅.

Let's calculate the probabilities of these intersections:

• P(A\(_1\) ∩ A\(_2\)) = $\frac{|A_1 \cap A_2|}{|\Omega|} = \frac{1}{4}$
• P(A\(_1\) ∩ A\(_3\)) = $\frac{|A_1 \cap A_3|}{|\Omega|} = \frac{1}{4}$
• P(A\(_2\) ∩ A\(_3\)) = $\frac{|A_2 \cap A_3|}{|\Omega|} = \frac{1}{4}$
• P(A\(_1\) ∩ A\(_2\) ∩ A\(_3\)) = $\frac{|A_1 \cap A_2 \cap A_3|}{|\Omega|} = \frac{0}{4} = 0$

Checking for Independence

Mutual Exclusion

Events are mutually exclusive if their intersection is empty. A\(_1\) ∩ A\(_2\) = {112}, which is not empty. Therefore, the events are not mutually exclusive.

Pairwise Independence

Two events A and B are pairwise independent if P(A ∩ B) = P(A) * P(B). We need to check this for all pairs of events (A\(_1\), A\(_2\)), (A\(_1\), A\(_3\)), and (A\(_2\), A\(_3\)).

• For A\(_1\) and A\(_2\):
  P(A\(_1\) ∩ A\(_2\)) = $\frac{1}{4}$
  P(A\(_1\)) * P(A\(_2\)) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
  Since P(A\(_1\) ∩ A\(_2\)) = P(A\(_1\)) * P(A\(_2\)), A\(_1\) and A\(_2\) are pairwise independent.
• For A\(_1\) and A\(_3\):
  P(A\(_1\) ∩ A\(_3\)) = $\frac{1}{4}$
  P(A\(_1\)) * P(A\(_3\)) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
  Since P(A\(_1\) ∩ A\(_3\)) = P(A\(_1\)) * P(A\(_3\)), A\(_1\) and A\(_3\) are pairwise independent.
• For A\(_2\) and A\(_3\):
  P(A\(_2\) ∩ A\(_3\)) = $\frac{1}{4}$
  P(A\(_2\)) * P(A\(_3\)) = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
  Since P(A\(_2\) ∩ A\(_3\)) = P(A\(_2\)) * P(A\(_3\)), A\(_2\) and A\(_3\) are pairwise independent.

Since all pairs of events are independent, the events A\(_1\), A\(_2\), and A\(_3\) are pairwise independent.

Mutual Independence (Independence)

Events A, B, and C are mutually independent if they are pairwise independent AND P(A ∩ B ∩ C) = P(A) * P(B) * P(C). We already checked pairwise independence. Now let's check the condition for the intersection of all three events.

• P(A\(_1\) ∩ A\(_2\) ∩ A\(_3\)) = 0
• P(A\(_1\)) * P(A\(_2\)) * P(A\(_3\)) = $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

Since P(A\(_1\) ∩ A\(_2\) ∩ A\(_3\)) ($\text{0}$) is not equal to P(A\(_1\)) * P(A\(_2\)) * P(A\(_3\)) ($\text{1/8}$), the events are not mutually independent.

Dependent Events

Events are dependent if they are not mutually independent. Since the events are not mutually independent (even though they are pairwise independent), they are considered dependent in the context of mutual independence, but the term "dependent events" usually implies *not* pairwise independent. The options provided distinguish between pairwise independent and independent (meaning mutually independent in this context). Since they satisfy the condition for pairwise independence but not mutual independence, "pairwise independent events" is the most specific and correct description among the choices.

Conclusion on Event Relationship

Based on our calculations:

• They are not mutually exclusive because intersections are non-empty.
• They are pairwise independent because P(A\(_i\) ∩ A\(_j\)) = P(A\(_i\)) * P(A\(_j\)) for all pairs.
• They are not mutually independent because P(A\(_1\) ∩ A\(_2\) ∩ A\(_3\)) ≠ P(A\(_1\)) * P(A\(_2\)) * P(A\(_3\)).

Therefore, the events A\(_1\), A\(_2\), and A\(_3\) are pairwise independent.

Evaluating the Options

• independent events: This usually refers to mutual independence. The events are not mutually independent.
• dependent events: This is a broad category. While they are not mutually independent, "pairwise independent" is a more precise description when that condition holds.
• pairwise independent events: This is true, as shown by our calculations for all pairs.
• mutually exclusive events: This is false, as intersections are non-empty.

The events A\(_1\), A\(_2\), and A\(_3\) are pairwise independent.

Revision Table: Key Probability Concepts

Additional Information on Event Independence

It's important to understand the difference between pairwise independence and mutual independence. Mutual independence is a stronger condition than pairwise independence. If a set of events is mutually independent, they are always pairwise independent. However, as demonstrated in this problem, if events are pairwise independent, they are not necessarily mutually independent.

In practical terms, pairwise independence means that knowing whether one event in a pair occurred does not change the probability of the other event in that pair occurring. Mutual independence means that knowing about any combination of other events (including single events or groups of events) does not change the probability of a specific event occurring.