The probability that a football team loosing a match at Kolkata is 3/5 and winning a match at Bengaluru is 6/7 ; the probability of the team winning at least one match is

Option c: 32/35. Solution: Probability of Winning At Least One Match Given: - P(lose at Kolkata) = 3/5 , so P(win at Kolkata) = 1 - 3/5 = 2/5 - P(win at Bengaluru) = 6/7 , so P(lose at Bengaluru) = 1/7 Using complement: P(win at least one) = 1 - P(lose both) = 1 - P(lose Kolkata) × P(lose Bengaluru) = 1 - 3/5 × 1/7 = 1 - 3/35 = 32/35 Hence, Option C is the correct answer.

The probability that a football team loosing a match at Kolkata is $3/5$ and winning a match at Bengaluru is $6/7$; the probability of the team winning at least one match is

The correct answer is Option c: $32/35$. **Probability of Winning At Least One Match** Given: - $P(\text{lose at Kolkata}) = \frac{3}{5}$, so $P(\text{win at Kolkata}) = 1 - \frac{3}{5} = \frac{2}{5}$ - $P(\text{win at Bengaluru}) = \frac{6}{7}$, so $P(\text{lose at Bengaluru}) = \frac{1}{7}$ Using complement: $$P(\text{win at least one}) = 1 - P(\text{lose both})$$ $$= 1 - P(\text{lose Kolkata}) \times P(\text{lose Bengaluru})$$ $$= 1 - \frac{3}{5} \times \frac{1}{7} = 1 - \frac{3}{35} = \frac{32}{35}$$ Hence, **Option C** is the correct answer.

ProbabilityMCQPYQ July 21Question 3327 of 187

All Questions

The probability that a football team loosing a match at Kolkata is $\displaystyle 3/5$ and winning a match at Bengaluru is $\displaystyle 6/7$ ; the probability of the team winning at least one match is

Options

\displaystyle 3/35

\displaystyle 18/35

\displaystyle 32/35

\displaystyle 17/35

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Correct Answer

✅ Option c — $\displaystyle 32/35$

All Options:

A $\displaystyle 3/35$
B $\displaystyle 18/35$
C $\displaystyle 32/35$
D $\displaystyle 17/35$

Detailed Solution & Explanation

**Probability of Winning At Least One Match** Given: -

\displaystyle P(\text{lose at Kolkata}) = \frac{3}{5}

, so

\displaystyle P(\text{win at Kolkata}) = 1 - \frac{3}{5} = \frac{2}{5}

\displaystyle P(\text{win at Bengaluru}) = \frac{6}{7}

, so

\displaystyle P(\text{lose at Bengaluru}) = \frac{1}{7}

Using complement:

P(\text{win at least one}) = 1 - P(\text{lose both})

= 1 - P(\text{lose Kolkata}) \times P(\text{lose Bengaluru})

= 1 - \frac{3}{5} \times \frac{1}{7} = 1 - \frac{3}{35} = \frac{32}{35}

Hence, **Option C** 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

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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The probability that a football team loosing a match at Kolkata is 3/5\displaystyle 3/53/5 and winning a match at Bengaluru is 6/7\displaystyle 6/76/7; the probability of the team winning at least one match is