Two bags I and II are given. Bag I contains 3 red and 5 black balls, while Bag II contains 4 red and 3 black b

Question

Q. Two bags I and II are given. Bag I contains 3 red and 5 black balls, while Bag II contains 4 red and 3 black balls. A ball is transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. If the ball drawn is black, find the probability that the transferred ball is black.

(OR)

An urn contains 5 red, 2 white and 3 black balls. Three balls are drawn at random from this urn one-by-one without replacement. Find the probability distribution of the number of white balls drawn. Also, find the mean and variance of the number of white balls drawn.

CBSECBSE Class XII Board 2020Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Conditional Probability

Conditional Probability

Roll a die and ask "what is the chance of an even number?" — that is $3/6$ . But suppose someone tells you the result is greater than 3. Now you are no longer looking at all six faces, only at $\{4,5,6\}$ , and two of those (4 and 6) are even, so the probability becomes $2/3$ . That change — from the probability of $A$ to the probability of $A$ given that $B$ has already occurred — is conditional probability.

The Idea: Shrink the Sample Space

Conditioning on $B$ throws away every outcome where $B$ is false and treats $B$ as the new "whole world." You measure $A$ only against what is still possible.

Note

Think of filtering a table of data: unconditional probability uses every row; conditional probability keeps only the rows where the condition is true.

The Definition

For events $A$ and $B$ with $P(B) > 0$ ,

P(A \mid B) = \frac{P(A \cap B)}{P(B)}.

We divide by $P(B)$ to rescale so that $B$ itself has probability $1$ ; the surviving part of $A$ is the overlap $A \cap B$ . Checking the die: $P(A \cap B) = P(\{4,6\}) = \tfrac{2}{6}$ and $P(B) = \tfrac{3}{6}$ , so $P(A \mid B) = \dfrac{2/6}{3/6} = \dfrac{2}{3}$ , matching the intuition.

Tip

Rearranging gives the multiplication rule $P(A \cap B) = P(A \mid B)\,P(B)$ , which is usually the easier way to compute a joint probability when a problem says "given that."

Two Cautions

$P(A \mid B)$ and $P(B \mid A)$ are generally not equal; swapping them is the classic mistake. They are linked by Bayes' theorem, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ . …

Part (b)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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$P(\text{transferred black}\mid\text{black drawn})=\dfrac{20}{29}$ .
Mean $=\dfrac35$ , Variance $=\dfrac{28}{75}$ .

Part (a)

Let $T_R,T_B$ be "red / black ball transferred": $P(T_R)=\tfrac38$ , $P(T_B)=\tfrac58$ . After the transfer Bag II holds $8$ balls.

If a red was transferred, Bag II $=5$ R, $3$ B $\Rightarrow P(\text{black}\mid T_R)=\tfrac38$ .
If a black was transferred, Bag II $=4$ R, $4$ B $\Rightarrow P(\text{black}\mid T_B)=\tfrac48=\tfrac12$ .

Total probability of drawing black:

P(B)=P(T_R)P(B|T_R)+P(T_B)P(B|T_B)=\frac38\cdot\frac38+\frac58\cdot\frac12=\frac{9}{64}+\frac{20}{64}=\frac{29}{64}.

By Bayes: …

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

Showing the 12 most recent of 94 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
The assertion is true: given the outcome is odd, the probability it is a prime is $\frac{2}{3}$ . The reason states the correct conditional probability formula. Since the reason directly justifies the calculation in the assertion, both are true and the reason is the correct explanation.

Concept first — Conditional probability asks: If we already know that event $B$ has occurred, what is the probability that event $A$ also occurs? The sample space shrinks from all possible outcomes to just those in $B$ . The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the precise way to compute this reduced probability.

Here, the die is unbiased, so each face $\{1,2,3,4,5,6\}$ has probability $\frac{1}{6}$ . The assertion involves two events:

$A$ : the number is prime. On a die, the primes are $2,3,5$ .

$B$ : the number is odd. The odd numbers are $1,3,5$ .

The condition "given that the number is odd" means we restrict attention to $B = \{1,3,5\}$ . Among these three equally likely outcomes, the primes are $3$ and $5$ — that's two out of three. So the conditional probability is $\frac{2}{3}$ .

Now let's verify step by step using the formula in Reason (R).

Define the events precisely.

$A = \{2,3,5\}$ , $B = \{1,3,5\}$ .

The sample space $S = \{1,2,3,4,5,6\}$ .

Compute $P(B)$ .

$B$ has 3 outcomes, each with probability $\frac{1}{6}$ , so $P(B) = \frac{3}{6} = \frac{1}{2}$ .

Compute $P(A \cap B)$ .

$A \cap B$ = numbers that are both prime and odd = $\{3,5\}$ . That's 2 outcomes, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$ .

Apply the formula from Reason (R).

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}.$
This matches the assertion exactly. …
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CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›Reveal solutionSolution

Since $S\cap F=F$ , the conditional probability $P(S\mid F)=1$ .

By the definition of conditional probability (with $P(F)\neq0$ ),

$P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$ …
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CBSE 2026Set CX1 markMCQ
$3P(A)=P(B)=\dfrac{5}{13}$
(a) $\dfrac{20}{39}$
(b) $\dfrac{16}{39}$
(c) $\dfrac{11}{39}$
(d) $\dfrac{14}{39}$
›Reveal solutionSolution

Using $P(A\cap B)=P(A/B)P(B)$ and the addition rule gives $P(A\cup B)=\dfrac{14}{39}$ — option (d).

Given: $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

So $P(B)=\dfrac{5}{13}$ and $P(A)=\dfrac{1}{3}\cdot\dfrac{5}{13}=\dfrac{5}{39}$ .

Intersection (multiplication rule):

$P(A\cap B)=P(A/B)\,P(B)=\dfrac{2}{5}\cdot\dfrac{5}{13}=\dfrac{2}{13}.$

…

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CBSE 2026Set A1 markMCQ
$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$
(a) $\frac{4}{9}$
(b) $\frac{4}{7}$
(c) $\frac{12}{13}$
(d) $\frac{1}{6}$
›Reveal solutionSolution

$P(A|B) = \frac{4}{9}$ .

Use the conditional-probability definition:

$P(A|B) = \frac{P(A\cap B)}{P(B)}.$

Substitute the given values: …

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CBSE 2026Set ANNUAL1 markMCQ
$P(B)=0.5$
(a) $\frac{15}{23}$
(b) $\frac{16}{25}$
(c) $\frac{16}{27}$
(d) $\frac{16}{23}$
›Reveal solutionSolution

By the definition of conditional probability, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{16}{25}$ .

The conditional probability of $A$ given $B$ is defined as

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B) eq0$

…

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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 2026Set ANNUAL1 mark
$Q. A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?$
›Reveal solutionSolution

List the equally likely outcomes for two children, restrict to those with at least one boy, then find the fraction that are both boys.

Sample space $=\{BB, BG, GB, GG\}$ , each equally likely.

Given at least one boy: reduced sample space $=\{BB,BG,GB\}$ (3 outcomes).

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.8$
(a) 0.8
(b) 0.5
(c) 0.32
(d) 0.4
›Reveal solutionSolution

Use the multiplication rule of conditional probability: $P(A\cap B) = P(B\mid A)\cdot P(A)$ .

Given $P(A)=0.8$ , $P(B\mid A)=0.4$ .

$P(A\cap B) = P(B\mid A)\cdot P(A) = 0.4 \times 0.8 = 0.32$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{3}{10}$
(a) $\dfrac{1}{4}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{5}{12}$
(d) $\dfrac{7}{12}$
›Reveal solutionSolution

Find $P(A\cap B)$ from the addition rule, then use the conditional probability formula.

Given $P(A)=\dfrac{3}{10}$ , $P(B)=\dfrac{2}{5}$ , $P(A\cup B)=\dfrac{3}{5}$ .

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{3}{10}, \dfrac{1}{5}, \dfrac{1}{10}$
›Reveal solutionSolution

The problem statement itself states that if the doctor comes by other means, he will never be late.

Let $E_1,E_2,E_3,E_4$ denote the events that the doctor comes by train, bus, scooter, or other means, with $P(E_1)=\dfrac{3}{10}$ , $P(E_2)=\dfrac{1}{5}$ , $P(E_3)=\dfrac{1}{10}$ , $P(E_4)=\dfrac{2}{5}$ (these sum to 1, confirming they form a complete set of cases).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P(A) = 1/2, P(B) = 3/8 and P(A\cupB) = 27/40 then P(A/B) is equal to:$
(a) 2/5
(b) 8/15
(c) 2/3
(d) 5/8
›Reveal solutionSolution

First find $P(A\cap B)$ using the addition rule, then apply the conditional probability formula $P(A/B)=\frac{P(A\cap B)}{P(B)}$ .

Given $P(A) = \frac12$ , $P(B)=\frac38$ , $P(A\cup B) = \frac{27}{40}$ .

Using the addition rule $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ :

$P(A\cap B) = P(A)+P(B)-P(A\cup B) = \frac{20}{40}+\frac{15}{40}-\frac{27}{40} = \frac{8}{40} = \frac15$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A\capB) = 0.32 Reason (R): Conditional Probability of 'B' when event A has occurred is given by P(B/A) = P(A\capB) / P(A)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Both Assertion (A) and Reason (R) are false.
›Reveal solutionSolution

Applying the conditional-probability formula in R directly to the given numbers reproduces the value claimed in A.

Reason (R): $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$ is the standard definition of conditional probability — true.

Checking Assertion (A): Given $P(A)=0.8,\ P(B/A)=0.4$ . Rearranging R's formula:

$P(A\cap B) = P(B/A)\times P(A) = 0.4 \times 0.8 = 0.32.$ …

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Part (a)

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