ProbabilityPYQ May 25Question 4381 of 187
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A father had three sons namely, Kailash, Harish and Prakash. All are above 65 years in age. Prakash happens to be the eldest while Kailash as youngest. As per the health history, it is estimated that the probability that Kailash survives another 5 years is 45\displaystyle \frac{4}{5}, Harish survives another 5 years is 35\displaystyle \frac{3}{5} and Prakash survives another 5 years is 12\displaystyle \frac{1}{2}. The probabilities that Kailash and Harish survive another 5 years is 0.46, Harish and Prakash survive another 5 years is 0.32 and Kailash and Prakash survive another 5 years is 0.48. The probability that all three sons survive another 5 years is 0.26. What shall be the probability that at least one of them survives another 5 years?

Options

A0.78
B0.72
C710\displaystyle \frac{7}{10}
D910\displaystyle \frac{9}{10}
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Correct Answer

Option d910\displaystyle \frac{9}{10}

All Options:

  • A0.78
  • B0.72
  • C710\displaystyle \frac{7}{10}
  • D910\displaystyle \frac{9}{10}

Detailed Solution & Explanation

Let us define the events for the survival of the three sons for another 5 years:
- K\displaystyle K: Kailash survives. P(K)=45=0.80\displaystyle P(K) = \frac{4}{5} = 0.80
- H\displaystyle H: Harish survives. P(H)=35=0.60\displaystyle P(H) = \frac{3}{5} = 0.60
- P\displaystyle P: Prakash survives. P(P)=12=0.50\displaystyle P(P) = \frac{1}{2} = 0.50

We are also given the joint probabilities:
- P(KH)=0.46\displaystyle P(K \cap H) = 0.46
- P(HP)=0.32\displaystyle P(H \cap P) = 0.32
- P(KP)=0.48\displaystyle P(K \cap P) = 0.48
- P(KHP)=0.26\displaystyle P(K \cap H \cap P) = 0.26

We want to find the probability that at least one of them survives, which is P(KHP)\displaystyle P(K \cup H \cup P). Using the addition theorem of probability for three events:
P(KHP)=P(K)+P(H)+P(P)P(KH)P(HP)P(KP)+P(KHP)P(K \cup H \cup P) = P(K) + P(H) + P(P) - P(K \cap H) - P(H \cap P) - P(K \cap P) + P(K \cap H \cap P) Substitute the given values into the formula:
P(KHP)=0.80+0.60+0.500.460.320.48+0.26P(K \cup H \cup P) = 0.80 + 0.60 + 0.50 - 0.46 - 0.32 - 0.48 + 0.26 P(KHP)=1.901.26+0.26P(K \cup H \cup P) = 1.90 - 1.26 + 0.26 P(KHP)=0.64+0.26=0.90P(K \cup H \cup P) = 0.64 + 0.26 = 0.90 Converting to fraction format:
0.90=9100.90 = \frac{9}{10}
Hence, **Option D** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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