Understanding Mutually Exclusive and Exhaustive Events

In probability, events are classified based on their relationship with each other. Two key concepts are mutually exclusive events and exhaustive events.

• Mutually Exclusive Events: These are events that cannot happen at the same time. If one event occurs, the other cannot. For example, when flipping a coin, getting heads and getting tails are mutually exclusive events. The probability of both occurring simultaneously is 0.
• Exhaustive Events: A set of events is exhaustive if at least one of the events must occur. In other words, the set of events covers all possible outcomes in the sample space. For example, when rolling a standard die, the events {getting a 1}, {getting a 2}, ..., {getting a 6} are exhaustive because one of these outcomes must happen.

If a set of events is both mutually exclusive and exhaustive, then the sum of their individual probabilities must equal 1. For events A, B, C, and D being mutually exclusive and exhaustive, we have:

\(P(A) + P(B) + P(C) + P(D) = 1\)

Given Probability Relationships

The problem provides a relationship between the probabilities of events A, B, C, and D:

\(2P(A) = 3P(B) = 4P(C) = 5P(D)\)

Expressing Probabilities in terms of a Constant

To solve this, we can set the common value of these equal expressions to a constant, say \(k\). So, we have:

\(2P(A) = 3P(B) = 4P(C) = 5P(D) = k\)

From this, we can express the probability of each event in terms of \(k\):

• From \(2P(A) = k\), we get \(P(A) = \frac{k}{2}\)
• From \(3P(B) = k\), we get \(P(B) = \frac{k}{3}\)
• From \(4P(C) = k\), we get \(P(C) = \frac{k}{4}\)
• From \(5P(D) = k\), we get \(P(D) = \frac{k}{5}\)

Using the Exhaustive Events Property to Solve for k

Since the events A, B, C, and D are exhaustive, the sum of their probabilities is 1:

\(P(A) + P(B) + P(C) + P(D) = 1\)

Substitute the expressions in terms of \(k\) into this equation:

\(\frac{k}{2} + \frac{k}{3} + \frac{k}{4} + \frac{k}{5} = 1\)

To sum these fractions, we find the least common multiple (LCM) of the denominators (2, 3, 4, 5). The LCM is 60. We multiply each term by 60:

\(60 \left( \frac{k}{2} \right) + 60 \left( \frac{k}{3} \right) + 60 \left( \frac{k}{4} \right) + 60 \left( \frac{k}{5} \right) = 60 \times 1\)

\(30k + 20k + 15k + 12k = 60\)

Combine the terms on the left side:

\((30 + 20 + 15 + 12)k = 60\)

\(77k = 60\)

Now, solve for \(k\):

\(k = \frac{60}{77}\)

Calculating P(A)

We know that \(P(A) = \frac{k}{2}\). Substitute the value of \(k\) we found:

\(P(A) = \frac{60/77}{2} = \frac{60}{77 \times 2}\)

\(P(A) = \frac{30}{77}\)

Final Calculation

The question asks for the value of \(77P(A)\). Substitute the value of \(P(A)\):

\(77P(A) = 77 \times \frac{30}{77}\)

The 77 in the numerator and denominator cancel out:

\(77P(A) = 30\)

Thus, \(77P(A)\) is equal to 30.

Revision Table: Key Concepts

Additional Information: Probability Basics

Probability is a measure of the likelihood of an event occurring. It is a value between 0 and 1, inclusive.

• A probability of 0 means the event is impossible.
• A probability of 1 means the event is certain.
• The sum of probabilities of all possible outcomes in a sample space is always 1.

When dealing with multiple events, understanding whether they are mutually exclusive, independent, exhaustive, etc., is crucial for applying the correct probability rules. In this problem, the mutually exclusive and exhaustive properties allowed us to set the sum of individual probabilities equal to 1, which was key to finding the unknown probabilities.