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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 AA, BB, CC and DD be mutually exclusive and exhaustive events and P(A)2=P(B)3=P(C)5=P(D)8\dfrac{P(A)}{2}=\dfrac{P(B)}{3}=\dfrac{P(C)}{5}=\dfrac{P(D)}{8}.

Question: What is P(A)+P(B)+P(C)P(A)+P(B)+P(C) equal to ?

Options

  1. A59\dfrac{5}{9}
  2. B118\dfrac{1}{18}
  3. C221\dfrac{2}{21}
  4. D713\dfrac{7}{13}

Correct answer

A. 59\dfrac{5}{9}

Step-by-step solution

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

P(A)=2kP(A) = 2k, P(B)=3kP(B) = 3k, P(C)=5kP(C) = 5k, P(D)=8kP(D) = 8k

Since AA, BB, CC and DD are mutually exclusive and exhaustive events, P(A)+P(B)+P(C)+P(D)=1P(A) + P(B) + P(C) + P(D) = 1

2k+3k+5k+8k=12k + 3k + 5k + 8k = 1

18k=1⇒k=11818k = 1 \Rightarrow k = \dfrac{1}{18}

P(A)+P(B)+P(C)=2k+3k+5k=10kP(A) + P(B) + P(C) = 2k + 3k + 5k = 10k

10k=10×118=5910k = 10 \times \dfrac{1}{18} = \dfrac{5}{9}

Answer: 59\dfrac{5}{9}
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