Passage: Let A , B , C and D be mutually exclusive and exhaustive events and (P(A))/(2)=(P(B))/(3)=(P(C))/(5)=(P(D))/(8) . Question: What is ([2P(A)+3P(B)])/([4P(C)+5P(D)]) equal to ?

Correct answer: B. (13)/(60) Let (P(A))/(2) = (P(B))/(3) = (P(C))/(5) = (P(D))/(8) = k Then P(A) = 2k , P(B) = 3k , P(C) = 5k , and P(D) = 8k Since A , B , C , and D are mutually exclusive and exhaustive events, P(A) + P(B) + P(C) + P(D) = 1 2k + 3k + 5k + 8k = 1 18k = 1 k = (1)/(18) We need to evaluate (2P(A) + 3P(B))/(4P(C) + 5P(D)) Substituting the values in terms of k : (2(2k) + 3(3k))/(4(5k) + 5(8k)) = (4k + 9k)/(20k + 40k) = (13k)/(60k) = (13)/(60) Answer: (13)/(60)

NDA 2026

Exam: NDA
Year: 2026
Subject: Mathematics
Chapter: Probability

NDA 2026 · Mathematics · Probability

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

Passage: Let

A

B

C

and

D

be mutually exclusive and exhaustive events and

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

.

Question: What is

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

equal to ?

Options

A $\dfrac{13}{18}$
B $\dfrac{13}{60}$
C $\dfrac{4}{21}$
D $\dfrac{5}{28}$

Correct answer

B. $\dfrac{13}{60}$

Step-by-step solution

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

, and

P(D) = 8k

Since

A

B

C

, and

D

are mutually exclusive and exhaustive events,

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

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

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

We need to evaluate

\dfrac{2P(A) + 3P(B)}{4P(C) + 5P(D)}

Substituting the values in terms of

k

\dfrac{2(2k) + 3(3k)}{4(5k) + 5(8k)} = \dfrac{4k + 9k}{20k + 40k} = \dfrac{13k}{60k} = \dfrac{13}{60}

Answer:

\dfrac{13}{60}

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