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Rupesh is known to hit a target in 5 out of 9 shots whereas David is known to hit the same target in 6 out of 11 shots. What is the probability that the target would be hit once they both try?

Options

A7999\displaystyle \frac{79}{99}
B1013\displaystyle \frac{10}{13}
C1426\displaystyle \frac{14}{26}
D1318\displaystyle \frac{13}{18}
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Correct Answer

Option a7999\displaystyle \frac{79}{99}

All Options:

  • A7999\displaystyle \frac{79}{99}
  • B1013\displaystyle \frac{10}{13}
  • C1426\displaystyle \frac{14}{26}
  • D1318\displaystyle \frac{13}{18}

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Detailed Solution & Explanation

Let R\displaystyle R represent the event that Rupesh hits the target, and D\displaystyle D represent the event that David hits the target. 1. The individual probabilities of hitting the target are: - P(R)=59\displaystyle P(R) = \frac{5}{9} - P(D)=611\displaystyle P(D) = \frac{6}{11} 2. The probabilities of failing to hit the target are: - P(R)=159=49\displaystyle P(R') = 1 - \frac{5}{9} = \frac{4}{9} - P(D)=1611=511\displaystyle P(D') = 1 - \frac{6}{11} = \frac{5}{11} 3. Assuming their shots are independent, the probability that neither of them hits the target is: P(Neither hits)=P(R)×P(D)=49×511=2099P(\text{Neither hits}) = P(R') \times P(D') = \frac{4}{9} \times \frac{5}{11} = \frac{20}{99} 4. The probability that the target is hit (at least one hits it) is: P(Target is hit)=1P(Neither hits)=12099=7999P(\text{Target is hit}) = 1 - P(\text{Neither hits}) = 1 - \frac{20}{99} = \frac{79}{99} Hence, **Option A** 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.

Key Concepts to Understand

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