Let A, B, C and D be mutually exclusive and exhaustive events and $\frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8}$.
Event Probability Calculation
The problem involves four events: A, B, C, and D. These events are described as mutually exclusive and exhaustive.
- Mutually exclusive means that only one of the events can occur at a time.
- Exhaustive means that one of these events must occur, covering all possibilities.
The relationship between their probabilities is given as:
\( \frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8} \)
Determining Probability Values
Let's represent this common ratio by a constant, \(k\):
\( k = \frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8} \)
From this, we can express the probability of each event in terms of \(k\):
- \( P(A) = 2k \)
- \( P(B) = 3k \)
- \( P(C) = 5k \)
- \( P(D) = 8k \)
Because the events A, B, C, and D are mutually exclusive and exhaustive, their total probability must equal 1:
\( P(A) + P(B) + P(C) + P(D) = 1 \)
Substitute the expressions in terms of \(k\) into the equation:
\( 2k + 3k + 5k + 8k = 1 \)
Summing the coefficients of \(k\):
\( (2 + 3 + 5 + 8)k = 1 \)
\( 18k = 1 \)
Solving for \(k\):
\( k = \frac{1}{18} \)
Calculating the Required Sum
The question asks for the value of \(P(A) + P(B) + P(C)\). Using the probabilities expressed in terms of \(k\):
\( P(A) + P(B) + P(C) = 2k + 3k + 5k \)
Summing these terms:
\( P(A) + P(B) + P(C) = (2 + 3 + 5)k \)
\( P(A) + P(B) + P(C) = 10k \)
Now substitute the value of \(k\) we found (\(k = \frac{1}{18}\)):
\( P(A) + P(B) + P(C) = 10 \times \frac{1}{18} \)
\( P(A) + P(B) + P(C) = \frac{10}{18} \)
Simplify the fraction:
\( P(A) + P(B) + P(C) = \frac{5}{9} \)
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