An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without repla

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Q. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls black?

Karnataka PUCKarnataka II PUC Board 2025Subjective· 2mImportance★★★★★

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Concept understanding — Hypergeometric Probability

Hypergeometric Probability: Drawing Without Replacement

You have a bag of 20 marbles: 12 red and 8 blue. You pick 5 without putting any back. What's the chance exactly 3 are red?

This is what the hypergeometric distribution handles. Unlike the binomial distribution, the trials are not independent — each draw changes the composition of the bag.

The Intuition

When you draw without replacement, the probability of a red on the second draw depends on the first: take a red first and fewer reds remain, so the next red is less likely. Hypergeometric probability captures exactly this dependency.

The setup: "I have a finite population split into two groups. I take a sample without replacement. What's the probability my sample has exactly $k$ items from the first group?"

The Precise Statement

P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

Where:

$N$ = total items in the population (20 marbles)
$K$ = number of "success" items ( $12$ red)
$n$ = number drawn (sample size, $5$ )
$k$ = successes wanted in the sample ( $3$ reds)

Why This Formula Makes Sense

The denominator $\binom{N}{n}$ counts all ways to choose $n$ items from $N$ — the equally likely outcomes. The numerator counts favourable ones:

$\binom{K}{k}$ : choose $k$ reds from the $K$ reds
$\binom{N-K}{n-k}$ : choose the remaining $n-k$ from the $N-K$ blues

Multiplying pairs each way of picking reds with each way of picking blues.

Worked Example

$N=20$ , $K=12$ , $n=5$ , $k=3$ :

P(\text{exactly 3 reds}) = \frac{\binom{12}{3} \binom{8}{2}}{\binom{20}{5}} = \frac{220 \times 28}{15504} = \frac{6160}{15504} \approx 0.397

About 39.7%.

Watch out

A common mistake is using the binomial formula here. Binomial assumes independent trials (drawing with replacement). With $p = 12/20 = 0.6$ it gives $\binom{5}{3}(0.6)^3(0.4)^2 \approx 0.346$ — close but wrong. The gap grows as the sample becomes a larger fraction of the population.

When to Use Hypergeometric …

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Draws without replacement — multiply $P(\text{1st black})\times P(\text{2nd black}\mid\text{1st black}) = \frac{10}{15}\times\frac{9}{14} = \frac{3}{7}$ .

The urn has $10$ black and $5$ white balls, a total of $15$ balls. Two balls are drawn one after the other without replacement.

Let

$B_1$ = the first ball drawn is black,
$B_2$ = the second ball drawn is black.

Step 1 — probability the first is black:

P(B_1) = \frac{10}{15}.

Step 2 — probability the second is black, given the first was black. After removing one black ball, $9$ black remain out of $14$ total: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Two cards are drawn at random without replacement from a pack of 52 playing cards, then the probability that both the cards are black in color, is$
(a) $1/2$
(b) $1/12$
(c) $25/102$
(d) $1/4$
›Reveal solutionSolution

Multiply the probability of the first card being black by the (conditional) probability of the second card being black given the first was black.

$P(\text{1st black}) = \dfrac{26}{52}=\dfrac12$ .

…

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CBSE 2024Set A11 markMCQ
$Q. If two cards are drawn without replacement from a pack of 52 playing cards then the probability that both the cards are black is$
(a) $\dfrac{1}{26}$
(b) $\dfrac{1}{4}$
(c) $\dfrac{25}{104}$
(d) $\dfrac{25}{102}$
›Reveal solutionSolution

Multiply the sequential probabilities of drawing black cards: $\tfrac{26}{52}\cdot\tfrac{25}{51}=\tfrac{25}{102}$ , so (d). …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Two cards are drawn at random and without replacement from a pack of 52 playing cards, then the probability that both the cards are black is:$
(a) $\dfrac{26}{52}$
(b) $\dfrac{52}{102}$
(c) $\dfrac{25}{51}$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

Multiply the probability of drawing a black card first by the probability of drawing a black card second (without replacement), since the events are dependent.

A standard deck has 26 black cards out of 52.

$P(\text{1st black}) = \dfrac{26}{52}$

After removing one black card (without replacement), 25 black cards remain out of 51 total:

$P(\text{2nd black} \mid \text{1st black}) = \dfrac{25}{51}$

$P(\text{both black}) = \dfrac{26}{52} \times \dfrac{25}{51} = \dfrac{650}{2652} = \dfrac{25}{102}$

This can also be checked with combinations: $\dfrac{\binom{26}{2}}{\binom{52}{2}} = \dfrac{325}{1326} = \dfrac{25}{102}$ .

…

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CBSE 2024Set ANNUAL1 mark
$Q. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball is ................ .$
›Reveal solutionSolution

$\dfrac{15}{56}$ .

Bag has $5$ red $+3$ blue $=8$ balls; $3$ are drawn without replacement.

Total ways to draw $3$ balls: $\binom{8}{3}=56$ .

…

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CBSE 2020Set 65/3/11 mark
$Q. Fill in the blank: A bag contains 3 black, 4 red and 2 green balls. If three balls are drawn simultaneously at random, then the probability that the balls are of different colours is __________.$
›Reveal solutionSolution
The key idea is to count the number of ways to pick one ball of each colour and divide by the total number of ways to draw any three balls. The probability is $\frac{12}{42} = \frac{2}{7}$ .

Concept & Intuition

When you draw balls “simultaneously,” order doesn’t matter — you’re just picking a set of three balls. The phrase “different colours” means you get exactly one black, one red, and one green. So the problem reduces to a classic “favourable outcomes / total outcomes” calculation using combinations.

The trap many students fall into is forgetting that the balls of the same colour are distinct objects (even though they look alike, each ball is a separate item). We treat them as distinct because we’re counting equally likely outcomes — each ball is equally likely to be chosen.

Step-by-step solution

Find the total number of balls.

Black: 3, Red: 4, Green: 2.

Total = $3 + 4 + 2 = 9$ balls.

Total number of ways to draw any 3 balls from 9.

Since order doesn’t matter, we use combinations:

$\text{Total outcomes} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.$

Count the favourable outcomes — one ball of each colour.

Choose 1 black from 3: $\binom{3}{1} = 3$ ways.

Choose 1 red from 4: $\binom{4}{1} = 4$ ways.

Choose 1 green from 2: $\binom{2}{1} = 2$ ways.

By the multiplication principle, the number of ways to pick one of each colour is:

$\text{Favourable outcomes} = 3 \times 4 \times 2 = 24.$

Compute the probability. …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. A card is drawn from a pack of 52 cards and then a second card is drawn without replacement. The probability that both cards drawn are queens is:$
(a) $\frac{1}{17}$
(b) $\frac{1}{221}$
(c) $\frac{1}{13}$
(d) None of these
›Reveal solutionSolution

Multiply the probabilities of drawing a queen first, then a queen from the remaining cards (dependent events, no replacement).

$P(\text{1st queen}) = \dfrac{4}{52}$ . Given the first card was a queen, $P(\text{2nd queen}) = \dfrac{3}{51}$ .

…

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Q.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls black?

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