ProbabilityMCQMTP Apr 21Question 2801 of 295
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If a card is drawn at random from a pack of cards, what is the chance of getting a Spade or an ace?

Options

A413\displaystyle \frac{4}{13}
B513\displaystyle \frac{5}{13}
C0.25\displaystyle 0.25
D0.20\displaystyle 0.20
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Correct Answer

✅ Option a — 413\displaystyle \frac{4}{13}

All Options:

  • A413\displaystyle \frac{4}{13}
  • B513\displaystyle \frac{5}{13}
  • C0.25\displaystyle 0.25
  • D0.20\displaystyle 0.20

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Detailed Solution & Explanation

Let S\displaystyle S represent the event of drawing a Spade and A\displaystyle A represent the event of drawing an Ace from a standard pack of 52 cards. 1. **Total number of cards** in a standard pack is n(Total)=52\displaystyle n(\text{Total}) = 52. 2. **Number of Spades** is n(S)=13\displaystyle n(S) = 13. Thus, the probability of drawing a Spade is: P(S)=1352P(S) = \frac{13}{52} 3. **Number of Aces** is n(A)=4\displaystyle n(A) = 4. Thus, the probability of drawing an Ace is: P(A)=452P(A) = \frac{4}{52} 4. **Number of cards that are both a Spade and an Ace** (the Ace of Spades) is n(S∩A)=1\displaystyle n(S \cap A) = 1. Thus, the probability of drawing the Ace of Spades is: P(S∩A)=152P(S \cap A) = \frac{1}{52} Using the addition theorem of probability, the probability of getting a Spade or an Ace is: P(S∪A)=P(S)+P(A)−P(S∩A)P(S \cup A) = P(S) + P(A) - P(S \cap A) P(S∪A)=1352+452−152=1652=413P(S \cup A) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} 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.

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