Determine Event Probabilities

Let the common ratio be k. According to the question:

• \( \frac{P(A)}{2} = \frac{P(B)}{3} = \frac{P(C)}{5} = \frac{P(D)}{8} = k \)

This implies:

• \( P(A) = 2k \)
• \( P(B) = 3k \)
• \( P(C) = 5k \)
• \( P(D) = 8k \)

Since events A, B, C, and D are mutually exclusive and exhaustive, their probabilities must sum to 1:

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

\( 2k + 3k + 5k + 8k = 1 \)

\( 18k = 1 \implies k = \frac{1}{18} \)

Now, substitute k back to find the probabilities:

• \( P(A) = 2 \times \frac{1}{18} = \frac{2}{18} \)
• \( P(B) = 3 \times \frac{1}{18} = \frac{3}{18} \)
• \( P(C) = 5 \times \frac{1}{18} = \frac{5}{18} \)
• \( P(D) = 8 \times \frac{1}{18} = \frac{8}{18} \)

Calculate Geometric Mean (G)

The geometric mean (G) of P(A), P(B), P(C), and P(D) is given by:

\( G = \left[ P(A) \cdot P(B) \cdot P(C) \cdot P(D) \right]^{\frac{1}{4}} \)

Substitute the calculated probabilities:

\( G = \left[ \frac{2}{18} \cdot \frac{3}{18} \cdot \frac{5}{18} \cdot \frac{8}{18} \right]^{\frac{1}{4}} \)

Multiply the numerators and denominators:

\( G = \left[ \frac{2 \cdot 3 \cdot 5 \cdot 8}{18^4} \right]^{\frac{1}{4}} \)

\( G = \left[ \frac{240}{18^4} \right]^{\frac{1}{4}} \)

Separate the terms:

\( G = \frac{(240)^{\frac{1}{4}}}{(18^4)^{\frac{1}{4}}} = \frac{(240)^{\frac{1}{4}}}{18} \)

Compute Final Value (9G)

We need to find the value of 9G:

\( 9G = 9 \times \frac{(240)^{\frac{1}{4}}}{18} \)

Simplify the expression:

\( 9G = \frac{(240)^{\frac{1}{4}}}{2} \)

To match the options, express 2 as a fourth root:

\( 2 = (2^4)^{\frac{1}{4}} = (16)^{\frac{1}{4}} \)

Substitute this back into the expression for 9G:

\( 9G = \frac{(240)^{\frac{1}{4}}}{(16)^{\frac{1}{4}}} \)

Combine the terms under the fourth root:

\( 9G = \left( \frac{240}{16} \right)^{\frac{1}{4}} \)

\( 9G = (15)^{\frac{1}{4}} \)