Q.4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Using classical probability, the number of favourable ways to draw 3 diamonds and 1 spade is , and the total ways to draw any 4 cards is . The required probability is .
Why classical probability works here
When a deck is well-shuffled, every set of 4 cards is equally likely. Classical probability says:
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.
A common mistake is to treat the draws as ordered (e.g., ). That overcounts because the same set of 4 cards can be drawn in 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:
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: ways
- 1 spade from the 13 available: ways
These two choices are independent — picking diamonds does not affect picking spades — so we multiply:
3. Write the probability expression.
4. Compute the combinations.
First, :
Next, .
So the numerator is:
Now :
Compute step by step:
Divide by :
So .
5. Simplify the fraction.
Both numerator and denominator are divisible by 13:
Thus:
Check if further simplification is possible. 286 factors as , and 20825 factors as — no common factors. So this is the simplest form.
You can also compute the probability using the multiplication rule for dependent events (without replacement):
The accounts for the spade appearing in any of the 4 positions. Try it — you'll get the same .
The probability of drawing 3 diamonds and 1 spade is .
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