CBSE 2026 · Region 5 · Set 1 · Q31 · 3 marks
A die is rolled. Consider events : $\displaystyle \mathrm{A}=\{1,2,5\}, \mathrm{B}=\{3,5\}, \mathrm{C}=\{2,3,4,5\}$ and hence find :(i)$\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{C})$ and $\displaystyle \mathrm{P}(\mathrm{C} \mid \mathrm{A})$(ii)$\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B} \mid \mathrm{C})$ and $\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B} \mid \mathrm{C})$A box contains $\displaystyle 6$ cards numbered $\displaystyle 1$ to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as $\displaystyle 10$, and B, the event of a number other than $\displaystyle 4$ on the first card selected. Find $\displaystyle \mathrm{P}(\mathrm{A}$ and B$\displaystyle )$ and find whether the events A and B are independent events or not.
A die is rolled. Consider events : $\displaystyle \mathrm{A}=\{1,2,5\}, \mathrm{B}=\{3,5\}, \mathrm{C}=\{2,3,4,5\}$ and hence find :
(i)
$\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{C})$ and $\displaystyle \mathrm{P}(\mathrm{C} \mid \mathrm{A})$
(ii)
$\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B} \mid \mathrm{C})$ and $\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B} \mid \mathrm{C})$
A box contains $\displaystyle 6$ cards numbered $\displaystyle 1$ to 6. A student is asked to pick up two cards, one by one after replacement and note down the numbers on the cards. Let A be the event of getting sum of the numbers on two cards as $\displaystyle 10$, and B, the event of a number other than $\displaystyle 4$ on the first card selected. Find $\displaystyle \mathrm{P}(\mathrm{A}$ and B$\displaystyle )$ and find whether the events A and B are independent events or not.
Marking-scheme solution
(a)
$\displaystyle P(A \mid C)=\dfrac{n(A \cap C)}{n(C)}=\dfrac{2}{4}$ or $\displaystyle \dfrac{1}{2}$
and $\displaystyle P(C \mid A)=\dfrac{n(C \cap A)}{n(A)}=\dfrac{2}{3}$
(ii)
$\displaystyle P(A \cap B \mid C)=\dfrac{n(A \cap B \cap C)}{n(C)}=\dfrac{1}{4}$
and $\displaystyle P(A \cup B \mid C)=\dfrac{n[(A \cup B) \cap C]}{n(C)}=\dfrac{3}{4}$
$\displaystyle A=\{(4,6),(6,4),(5,5)\}$ or $\displaystyle n(A)=3$
$\displaystyle n(B)=30$
$\displaystyle n(A \cap B)=2$
$\displaystyle P(A \cap B)=\dfrac{2}{36}$
$\displaystyle P(A)=\dfrac{3}{36}=\dfrac{1}{12}$
$\displaystyle P(B)=\dfrac{30}{36}=\dfrac{5}{6}$
$\displaystyle P(A) \cdot P(B)=\dfrac{5}{72} \neq P(A \cap B)$
Hence A and B are not independent
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.