What is {[2 {P}({A})+3 {P}({B})]}{[4 {P}({C})+5 {P}({D})]} equal to ?

Correct Answer is : 136013/606013

Correct Answer is : 136013/606013

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 : 107 of 120

Marks: +1, -0

What is

\frac{[2 \mathrm{P}(\mathrm{A})+3 \mathrm{P}(\mathrm{B})]}{[4 \mathrm{P}(\mathrm{C})+5 \mathrm{P}(\mathrm{D})]}

equal to ?

Solution:

Concept:

Use a common ratio to express all probabilities, then use the total probability = 1.

Explanation:

Let

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

So,

P(A) = 2k

P(B) = 3k

P(C) = 5k

P(D) = 8k

Since A, B, C, D are exhaustive events,

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

2k + 3k + 5k + 8k = 1 \Rightarrow 18k = 1 \Rightarrow k = \frac{1}{18}

Now, evaluating the required expression:

\frac{2P(A) + 3P(B)}{4P(C) + 5P(D)} = \frac{2(2k) + 3(3k)}{4(5k) + 5(8k)}

= \frac{4k + 9k}{20k + 40k} = \frac{13k}{60k} = \frac{13}{60}

Answer:

\frac{13}{60}

The correct option is .

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