Q.Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Concept: Conditional Probability — we restrict the sample space to the condition given.

Step 1: Total cards: 1 to 10. Condition: number > 3. So the reduced sample space is $\{4,5,6,7,8,9,10\}$ — 7 equally likely outcomes.

Step 2: Among these, the even numbers are $\{4,6,8,10\}$ — 4 favourable outcomes.

Step 3: Required probability = $\frac{\text{favourable outcomes in condition}}{\text{total outcomes in condition}} = \frac{4}{7}$.

The key here is conditional probability — we are not finding the probability of drawing an even card from all ten cards. Instead, we already know that the card shows a number greater than 3. That extra information shrinks the set of possible outcomes. The question becomes: Out of the cards that are >3, what fraction are even?

Let’s walk through it step by step.

1. Original sample space

   The cards are numbered 1 through 10. So the total number of equally likely outcomes when drawing one card is 10.

2. The condition: number > 3

   The cards that satisfy “more than 3” are:

That’s 7 cards. This becomes our reduced sample space — we only consider these 7 outcomes.

3. Favourable outcomes: even numbers among these From the set above, the even numbers are:

That’s 4 cards.

4. Apply the conditional probability formula If $A$ is the event “card is even” and $B$ is the event “card > 3”, then

Here $A \cap B$ = “even and >3” = $\{4,6,8,10\}$, so $P(A \cap B) = \frac{4}{10}$.

And $P(B) = \frac{7}{10}$.

Therefore

Method: Reduced-sample-space counting under a numeric condition

Use this when a condition restricts an equally likely draw to a sub-range (e.g. "the number is more than 3") and you want the chance of a further property within that range.

Steps

Step 1: Restrict the sample space to what the condition allows.

From the full list of equally likely outcomes, keep only those satisfying the "given" condition. This restricted list becomes the denominator count.

Step 2: Count how many of the surviving outcomes are favourable.

Within the restricted list, count the outcomes that also satisfy the event you want.

Step 3: Divide directly.

Equivalently $P(A\cap B)/P(B)$ with a common denominator that cancels. The key discipline is never to divide by the original total — always by the size of the restricted space.

Common Mistakes

Mistake 1: Dividing by the original total of $10$ instead of the restricted count.

Why it's wrong: once we know the number is $>3$, only $7$ cards $\{4,\dots,10\}$ remain possible, so the denominator is $7$, not $10$. Correct approach: use the reduced sample space, giving $\tfrac47$, not $\tfrac{4}{10}$.

Mistake 2: Miscounting the evens in the restricted range.

Why it's wrong: the evens greater than $3$ are $\{4,6,8,10\}$ — four of them; forgetting $10$ or including $2$ changes the numerator. Correct approach: list the restricted set explicitly before counting favourable outcomes.