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Miscellaneous Exercise · Q2

Q.4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

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Using classical probability, the number of favourable ways to draw 3 diamonds and 1 spade is (133)×(131)\binom{13}{3} \times \binom{13}{1}, and the total ways to draw any 4 cards is (524)\binom{52}{4}. The required probability is 28620825\frac{286}{20825}.

Why classical probability works here

When a deck is well-shuffled, every set of 4 cards is equally likely. Classical probability says:

P(event)=number of favourable outcomestotal number of equally likely outcomesP(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

The key is counting correctly. We are drawing without replacement and the order of the cards does not matter — only which 4 cards end up in your hand. So we use combinations, not permutations.

Watch out

A common mistake is to treat the draws as ordered (e.g., 52×51×50×4952 \times 51 \times 50 \times 49). That overcounts because the same set of 4 cards can be drawn in 4!4! different orders. Always ask: does order matter? Here it does not — we just want a set of cards.


Step-by-step solution

1. Count the total number of ways to draw any 4 cards from 52.

Since order does not matter, we use the combination formula:

Total outcomes=(524)\text{Total outcomes} = \binom{52}{4}

This is the denominator of our probability.

2. Count the favourable outcomes: exactly 3 diamonds and 1 spade.

A standard deck has 13 diamonds and 13 spades. We need to choose:

  • 3 diamonds from the 13 available: (133)\binom{13}{3} ways
  • 1 spade from the 13 available: (131)\binom{13}{1} ways

These two choices are independent — picking diamonds does not affect picking spades — so we multiply:

Favourable outcomes=(133)×(131)\text{Favourable outcomes} = \binom{13}{3} \times \binom{13}{1}

3. Write the probability expression.

P=(133)×(131)(524)P = \frac{\binom{13}{3} \times \binom{13}{1}}{\binom{52}{4}}

4. Compute the combinations.

First, (133)\binom{13}{3}:

(133)=13×12×113×2×1=17166=286\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = \frac{1716}{6} = 286

Next, (131)=13\binom{13}{1} = 13.

So the numerator is:

286×13=3718286 \times 13 = 3718

Now (524)\binom{52}{4}:

(524)=52×51×50×494×3×2×1\binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1}

Compute step by step:

  • 52×51=265252 \times 51 = 2652
  • 2652×50=1326002652 \times 50 = 132600
  • 132600×49=6497400132600 \times 49 = 6497400

Divide by 2424:

649740024=270725\frac{6497400}{24} = 270725

So (524)=270725\binom{52}{4} = 270725.

5. Simplify the fraction.

P=3718270725P = \frac{3718}{270725}

Both numerator and denominator are divisible by 13:

  • 3718÷13=2863718 \div 13 = 286
  • 270725÷13=20825270725 \div 13 = 20825

Thus:

P=28620825P = \frac{286}{20825}

Check if further simplification is possible. 286 factors as 2×11×132 \times 11 \times 13, and 20825 factors as 52×72×175^2 \times 7^2 \times 17 — no common factors. So this is the simplest form.

Tip

You can also compute the probability using the multiplication rule for dependent events (without replacement):

P=1352×1251×1150×1349×(41)P = \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50} \times \frac{13}{49} \times \binom{4}{1}

The (41)\binom{4}{1} accounts for the spade appearing in any of the 4 positions. Try it — you'll get the same 28620825\frac{286}{20825}.


✓Final answer

The probability of drawing 3 diamonds and 1 spade is 28620825\boxed{\frac{286}{20825}}.

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