If G is the geometric mean of {P}({A}), {P}({B}), {P}({C}) and {P}({D}), then what is 9 G equal to ?

Correct Answer is : 151415¹/4}1541

Correct Answer is : 151415¹/4}1541

Test Index

UPSC NDA I 2026 Mathematics Paper

Question Numbers: 106-108

For the next three (03) items that follow:

Let A, B, C and D be mutually exclusive and exhaustive events and

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

Question : 108 of 120

Marks: +1, -0

\mathrm{P}(\mathrm{A})

\mathrm{P}(\mathrm{B}), \mathrm{P}(\mathrm{C})

\mathrm{P}(\mathrm{D})

, then what is 9 G equal to ?

Solution:

Concept:

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

1

Use the given proportionality to find each probability, then compute the geometric mean.

Explanation:

Let

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

Then

\mathrm{P(A)}=2k

\mathrm{P(B)}=3k

\mathrm{P(C)}=5k

\mathrm{P(D)}=8k

Since they are exhaustive,

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

18k=1

, giving

k=\frac{1}{18}

Thus,

\mathrm{P(A)}=\frac{1}{9}

\mathrm{P(B)}=\frac{1}{6}

\mathrm{P(C)}=\frac{5}{18}

\mathrm{P(D)}=\frac{4}{9}

Geometric mean:

G=\left(\frac{1}{9}\cdot\frac{1}{6}\cdot\frac{5}{18}\cdot\frac{4}{9}\right)^{1/4}=\left(\frac{5}{2187}\right)^{1/4}

Now compute

9G

9G=9\left(\frac{5}{2187}\right)^{1/4}=3^2\cdot\frac{5^{1/4}}{3^{7/4}}=3^{1/4}\cdot5^{1/4}=15^{1/4}

Answer:

15^{\frac{1}{4}}

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