Problem 83 - Applied Statistics and Probability for Engineers

Question

\(+\) If \(A, B,\) and \(C\) are mutually exclusive events with \(P(A)=0.2, P(B)=0.3,\) and \(P(C)=0.4,\) determine the follow. ing probabilities: (a) \(P(A \cup B \cup C)\) (b) \(P(A \cap B \cap C)\) (c) \(P(A \cap B)\) (d) \(P[(A \cup B) \cap C]\) (e) \(P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)\)

Step-by-Step Solution

Answer

(a) 0.9, (b) 0, (c) 0, (d) 0, (e) 0.1

Step 1: Understand Mutually Exclusive Events

Mutually exclusive events are events that cannot happen at the same time. This means if one event occurs, the others cannot. Mathematically, for mutually exclusive events, such as events A, B, and C: \( P(A \cap B) = P(B \cap C) = P(A \cap C) = P(A \cap B \cap C) = 0 \).

Step 2: Calculate \(P(A \cup B \cup C)\)

According to the principle of mutually exclusive events, the probability of the union of mutually exclusive events is simply the sum of their probabilities. Therefore, \( P(A \cup B \cup C) = P(A) + P(B) + P(C) \). Substituting the given probabilities, we get \( P(A \cup B \cup C) = 0.2 + 0.3 + 0.4 = 0.9 \).

Step 3: Calculate \(P(A \cap B \cap C)\)

As mentioned earlier, since events A, B, and C are mutually exclusive, \( P(A \cap B \cap C) = 0 \).

Step 4: Calculate \(P(A \cap B)\)

For mutually exclusive events, \( P(A \cap B) = 0 \). This is because if event A occurs, event B cannot occur simultaneously.

Step 5: Calculate \(P[(A \cup B) \cap C]\)

Since events A, B, and C are mutually exclusive, any intersection of A and B with C results in an empty event (zero probability). Therefore, \( P[(A \cup B) \cap C] = P(A \cap C) + P(B \cap C) = 0 + 0 = 0 \).

Step 6: Calculate \(P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right)\)

First, compute \( P(A \cup B \cup C) \) which we found to be 0.9 in Step 2. Then, its complement \( A', B', \text{and} C' \) means that none of A, B, or C occurs, which gives us \( P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) = 1 - P(A \cup B \cup C) = 1 - 0.9 = 0.1 \).

Key Concepts

Mutually Exclusive Events

Mutually exclusive events are foundational elements in probability theory that simplify many calculations. These are events that cannot occur simultaneously. For example, if you flip a coin, landing on "heads" and "tails" are mutually exclusive events, because the coin cannot show both sides at once.

In mathematical terms, if events A, B, and C are mutually exclusive, as in our problem, then:

• The probability of both events A and B happening simultaneously is zero: \( P(A \cap B) = 0 \).
• The probability of all three events happening at the same time is also zero: \( P(A \cap B \cap C) = 0 \).

This makes it easier to calculate probabilities related to combinations of these events, because you only consider the individual probabilities of the events themselves.

Probability of Union

The probability of the union of mutually exclusive events is straightforward to calculate. It is the sum of the probabilities of each individual event. The union of events is when any one of the events occurs. In our example with events A, B, and C, the probability that at least one of these events happens is given by \( P(A \cup B \cup C) \).

Because events A, B, and C are mutually exclusive:

• You simply add their probabilities: \( P(A \cup B \cup C) = P(A) + P(B) + P(C) \).
• Using the values given, we have \( P(A \cup B \cup C) = 0.2 + 0.3 + 0.4 = 0.9 \).

This means there is a 90% chance that at least one of these events will occur.

Complementary Probability

The concept of complementary probability is a key idea in probability theory. It refers to the likelihood that an event does not occur. Symbolized by the prime notation (\(A'\) for "not A"), it represents all the outcomes in the sample space that are not part of the event.

The complementary probability is found using:

• \( P(A') = 1 - P(A) \), where 1 represents the certainty of all possible outcomes combined.
• In the given exercise, we needed to find \( P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) \), or the probability that none of the events A, B, or C occurs.

This can be determined by recognizing that this is the complement of the union of A, B, or C:
\( P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) = 1 - P(A \cup B \cup C) \)
Since we calculated \( P(A \cup B \cup C) = 0.9 \), the probability that none of the events occurs is:
\( P\left(A^{\prime} \cap B^{\prime} \cap C^{\prime}\right) = 1 - 0.9 = 0.1 \).
This tells us that there is a 10% chance that none of the events will happen.