Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots. One day Ar

Question

Q. Arka bought two cages of birds: Cage-I contains 5 parrots and 1 owl and Cage-II contains 6 parrots. One day Arka forgot to lock both cages and two birds flew from Cage-I to Cage-II (simultaneously). Then two birds flew back from Cage-II to Cage-I (simultaneously). Assume that all the birds have equal chances of flying. On the basis of the above information, answer the following questions:

(i) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, then find the probability that the owl is still in Cage-I. [2 Marks]

(ii) When two birds flew from Cage-I to Cage-II and two birds flew back from Cage-II to Cage-I, the owl is still seen in Cage-I, what is the probability that one parrot and the owl flew from Cage-I to Cage-II? [2 Marks]

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✓ Free question

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, $P(A \mid B) = \dfrac{P(B \mid A)\,P(A)}{P(B)}$ .
If $P(A \mid B) = P(A)$ , then knowing $B$ tells you nothing about $A$ — the events are independent. That is a special case, not the general rule.

Conditional probability is the foundation of the multiplication theorem, independence, and Bayes' theorem — every "given that" question in this chapter rests on it.

Conditional Probability opens the CBSE Class 12 Probability chapter and is foundational for everything that follows in that unit, including Bayes' theorem and the multiplication rule — making "conditional probability formula class 12 with examples" one of the most searched topics in Class 12 Mathematics. It is equally important for JEE Main and CUET, where conditional probability questions are set almost every year.

Concept: Conditional Probability & Bayes' Theorem

We track the owl through two transfers. Let event $A$ = "owl is still in Cage-I after both transfers". For part (ii), we need $P(\text{one parrot + owl left Cage-I} \mid A)$ .

(i) Probability that the owl is still in Cage-I

Step 1 – First transfer (Cage-I → Cage-II):

Cage-I has 5 parrots (P) and 1 owl (O). Two birds fly out.

Owl stays in Cage-I only if it is not among the two that left.
Number of ways to choose 2 birds from 6: $\binom{6}{2} = 15$
Ways to choose 2 parrots (no owl): $\binom{5}{2} = 10$ So $P(\text{owl stays after first transfer}) = \frac{10}{15} = \frac{2}{3}$ .

Step 2 – Second transfer (Cage-II → Cage-I):

After first transfer, Cage-II has 6 original parrots + 2 new birds.

If owl stayed in Cage-I, Cage-II has no owl — only 8 parrots.
Two birds fly back from Cage-II to Cage-I. The owl is safe regardless of which two return. So $P(\text{owl stays after second transfer} \mid \text{owl stayed after first}) = 1$ .

Step 3 – Final probability:

$P(A) = \frac{2}{3} \times 1 = \frac{2}{3}$ .

✓Final answer

\boxed{\frac{2}{3}}

(ii) Probability that one parrot and the owl left Cage-I, given the owl is still in Cage-I

Let event $B$ = "one parrot and the owl flew from Cage-I to Cage-II".

Step 1 – Find $P(B)$ :

From Cage-I, choose 1 parrot and the owl: $\binom{5}{1} \times \binom{1}{1} = 5$ ways. Total ways = 15.

So $P(B) = \frac{5}{15} = \frac{1}{3}$ .

Step 2 – Find $P(A \mid B)$ :

If $B$ occurs, the owl goes to Cage-II. For it to return to Cage-I, it must be among the two birds that fly back.

After first transfer, Cage-II has 6 parrots + (1 parrot + 1 owl) = 8 birds.

Ways to choose 2 birds from 8: $\binom{8}{2} = 28$
Ways to choose the owl + 1 other: $\binom{1}{1} \times \binom{7}{1} = 7$ So $P(\text{owl returns} \mid B) = \frac{7}{28} = \frac{1}{4}$ . Thus $P(A \mid B) = \frac{1}{4}$ .

Step 3 – Apply Bayes' Theorem:

$P(B \mid A) = \frac{P(A \mid B) \cdot P(B)}{P(A)} = \frac{\frac{1}{4} \times \frac{1}{3}}{\frac{2}{3}} = \frac{1/12}{2/3} = \frac{1}{8}$ .

✓Final answer

\boxed{\frac{1}{8}}

(i) $P(\text{owl still in Cage-I}) = \dfrac{3}{4}$ . (ii) Given the owl is still in Cage-I, $P(\text{one parrot and the owl flew out}) = \dfrac{1}{9}$ .

Cage-I starts with 5 parrots + 1 owl (6 birds); Cage-II has 6 parrots. Two birds go I $\to$ II, then two of the resulting 8 birds in Cage-II go II $\to$ I.

Part (i). The owl ends in Cage-I in two mutually exclusive ways.

Owl never leaves (not among the 2 that flew out of Cage-I):

P = \frac{\binom{5}{2}}{\binom{6}{2}} = \frac{10}{15} = \frac{2}{3}.

If the owl stays, it remains in Cage-I regardless of the return flight.

Owl flies out, then flies back. It flies out with

P = \frac{\binom{5}{1}}{\binom{6}{2}} = \frac{5}{15} = \frac{1}{3}.

Cage-II now holds 8 birds (incl. the owl); the owl is among the 2 that return with

P = \frac{\binom{7}{1}}{\binom{8}{2}} = \frac{7}{28} = \frac{1}{4}.

So this path contributes $\tfrac{1}{3}\cdot\tfrac{1}{4} = \tfrac{1}{12}$ .

P(\text{owl in Cage-I}) = \frac{2}{3} + \frac{1}{12} = \frac{8}{12} + \frac{1}{12} = \frac{9}{12} = \frac{3}{4}.

Part (ii). "One parrot and the owl flew out" is exactly the owl-flies-out case, and it is conditioned on the owl ending in Cage-I. Using Bayes' rule:

P(\text{owl flew out} \mid \text{owl in Cage-I}) = \frac{P(\text{owl flew out and returned})}{P(\text{owl in Cage-I})} = \frac{\tfrac{1}{12}}{\tfrac{3}{4}} = \frac{1}{12}\cdot\frac{4}{3} = \frac{1}{9}.

✓Final answer

\dfrac{3}{4}

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 88 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.

Watch out

A common mistake is to forget that the condition reduces the sample space. Some students count primes among all six numbers ( $3$ primes) and odd numbers among all six ( $3$ odds), then incorrectly write $\frac{3}{3}=1$ or $\frac{3}{6} \div \frac{3}{6}=1$ . The formula forces you to consider only the overlap, which correctly gives $\frac{2}{3}$ .

Check the truth of Reason (R).

The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the standard definition of conditional probability (provided $P(B) \neq 0$ ). It is always true. So Reason (R) is true.

Does Reason (R) correctly explain Assertion (A)?

Yes — the assertion's value $\frac{2}{3}$ is obtained directly by substituting the probabilities into this formula. The reason is not just a true statement; it is the very tool used to verify the assertion.

Tip

When both assertion and reason are true, and the reason is the principle that justifies the assertion, the answer is option (A). If the reason were true but irrelevant to the assertion, it would be option (B).

✓Final answer
$The correct option is (A) — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).$
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 $P(F)\neq0$ ),
$P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$
The sample space $S$ contains every outcome, so $S\cap F=F$ and $P(S\cap F)=P(F)$ . Therefore
$P(S\mid F)=\frac{P(F)}{P(F)}=1.$
✓Final answer
$1$
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):

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

Union (addition rule), with common denominator $39$ :

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac{5}{39}+\dfrac{15}{39}-\dfrac{6}{39}=\dfrac{14}{39}.$

✓Final answer
$\;P(A\cup B)=\dfrac{14}{39}$
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:

$P(A|B) = \frac{\frac{4}{13}}{\frac{9}{13}} = \frac{4}{13}\cdot\frac{13}{9} = \frac{4}{9}.$

✓Final answer
$\frac{4}{9}$
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

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

Substituting the given values $P(B)=0.5$ and $P(A\cap B)=0.32$ :

$P(A\mid B)=\frac{0.32}{0.5}=0.64=\frac{64}{100}=\frac{16}{25}$

✓Final answer
$\frac{16}{25}$
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).

Of these, both boys occurs only in $\{BB\}$ (1 outcome).

$P(\text{both boys}\mid\text{at least one boy}) = \dfrac13$ .

✓Final answer
$1/3$
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 $P(A)=0.8$ , $P(B\mid A)=0.4$ .

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

(Note: $P(B)=0.5$ is extra information not needed for this computation.)

✓Final answer
$0.32$
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 $P(A)=\dfrac{3}{10}$ , $P(B)=\dfrac{2}{5}$ , $P(A\cup B)=\dfrac{3}{5}$ .

$P(A\cap B)=P(A)+P(B)-P(A\cup B)=\dfrac{3}{10}+\dfrac{4}{10}-\dfrac{6}{10}=\dfrac{1}{10}$

$P(B/A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1/10}{3/10}=\dfrac{1}{3}$

✓Final answer
$\dfrac{1}{3}$
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).

Let $L$ be the event that he is late. We are given $P(L|E_1)=\dfrac{1}{4}$ , $P(L|E_2)=\dfrac{1}{3}$ , $P(L|E_3)=\dfrac{1}{12}$ , and "if he comes by other means, he will not be late", i.e. $P(L|E_4)=0$ .

So $P(\text{not late}\,|\,E_4)=1-P(L|E_4)=1-0=1$ .

✓Final answer
$P(\text{not late}\mid \text{other means}) = 1$
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 $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ :

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

Now, conditional probability:

$P(A/B) = \frac{P(A\cap B)}{P(B)} = \frac{1/5}{3/8} = \frac15\times\frac83 = \frac{8}{15}$

✓Final answer
$P(A/B) = \dfrac{8}{15}$
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:

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

This matches A exactly, so A is true, and R is precisely the formula used to derive it.

(Note: $P(B)=0.5$ is extra information, not needed for this particular computation.)

✓Final answer
$Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A. (Option a)$
CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{1}{2}$
(a) 0
(b) $\dfrac{1}{2}$
(c) 1
(d) Not defined
›Reveal solutionSolution

Conditional probability is defined as $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ , which requires dividing by $P(B)$ — impossible when $P(B)=0$ .

Given $P(A)=\dfrac12$ , $P(B)=0$ .

By definition:

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

Since $P(B)=0$ , this involves division by zero, which is undefined regardless of the value of $P(A\cap B)$ (and note $P(A\cap B)\le P(B)=0$ forces $P(A\cap B)=0$ too, giving the indeterminate $\frac00$ ).

$P(A\mid B) \text{ is not defined.}$

✓Final answer
$(d) Not defined$

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