Passage: Let A , B , C and D be mutually exclusive and exhaustive...

Since $A$, $B$, $C$ and $D$ are mutually exclusive and exhaustive events, the sum of their probabilities is $1$.

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

Let $\dfrac{P(A)}{2} = \dfrac{P(B)}{3} = \dfrac{P(C)}{5} = \dfrac{P(D)}{8} = k$.

Then $P(A) = 2k$, $P(B) = 3k$, $P(C) = 5k$, $P(D) = 8k$.

Substituting these into the sum equation gives:

$2k + 3k + 5k + 8k = 1$

$18k = 1 \Rightarrow k = \dfrac{1}{18}$

The probabilities are $P(A) = \dfrac{2}{18}$, $P(B) = \dfrac{3}{18}$, $P(C) = \dfrac{5}{18}$, $P(D) = \dfrac{8}{18}$.

The geometric mean $G$ of the four probabilities is:

$G = (P(A) P(B) P(C) P(D))^{\dfrac{1}{4}}$

$G = \left( \dfrac{2}{18} \times \dfrac{3}{18} \times \dfrac{5}{18} \times \dfrac{8}{18} \right)^{\dfrac{1}{4}}$

$G = \left( \dfrac{240}{18^4} \right)^{\dfrac{1}{4}}$

Simplifying the expression inside the bracket:

$G = \left( \dfrac{15 \times 16}{16 \times 3^8} \right)^{\dfrac{1}{4}} = \left( \dfrac{15}{3^8} \right)^{\dfrac{1}{4}}$

$G = \dfrac{15^{\dfrac{1}{4}}}{3^2} = \dfrac{15^{\dfrac{1}{4}}}{9}$

Multiplying by $9$ yields:

$9G = 15^{\dfrac{1}{4}}$

Answer: $15^{\frac{1}{4}}$