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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 4 · Set 1 · Q36
Some students are having a misconception while comparing decimals. For example, a student may mention that 78 · 56>78 · 9 as 7856>789. In order to…
Some students are having a misconception while comparing decimals. For example, a student may mention that $\displaystyle 78 \cdot 56>78 \cdot 9$ as $\displaystyle 7856>789$. In order to assess this concept, a decimal comparison test was administered to the students of class VI through the following question : In the recently held Sports Day in the school, $\displaystyle 5$ students participated in a javelin throw competition. The distances to which they have thrown the javelin are shown below in the table :
The students were asked to identify who has thrown the javelin the farthest. Based on the test attempted by the students, the teacher concludes that $\displaystyle 40 \%$ of the students have the misconception in the concept of decimal comparison and the rest do not have the misconception. $\displaystyle 80 \%$ of the students having misconception answered Bijoy as the correct answer in the paper. $\displaystyle 90 \%$ of the students who are identified with not having misconception, did not answer Bijoy as their answer. On the basis of the above information, answer the following questions :
| Name of student | Distance of javelin (in meters) |
| Ajay | $\displaystyle 47 \cdot 7$ |
| Bijoy | $\displaystyle 47 \cdot 07$ |
| Kartik | $\displaystyle 43 \cdot 09$ |
| Dinesh | $\displaystyle 43 \cdot 9$ |
| Devesh | $\displaystyle 45 \cdot 2$ |
(i)
What is the probability of a student not having misconception but still answers Bijoy in the test?
(ii)
What is the probability that a randomly selected student answers Bijoy as his answer in the test?
(iii)
(a)
What is the probability that a student who answered as Bijoy is having misconception?
OR
(b)
What is the probability that a student who answered as Bijoy is amongst students who do not have the misconception? Case Study - $\displaystyle 2$
Marking-scheme solution
Let $\displaystyle E_1$ : Student has a misconception
$\displaystyle E_2$ : Student does not have misconception
A : Student answered Bijoy as correct
$\displaystyle \therefore P(E_1)=\dfrac{40}{100},\ P(E_2)=\dfrac{60}{100}$
$\displaystyle P(A\mid E_1)=\dfrac{80}{100},\ P(A\mid E_2)=\dfrac{10}{100}$
$\displaystyle P(\bar A\mid E_1)=\dfrac{20}{100},\ P(\bar A\mid E_2)=\dfrac{90}{100}$
(i)
$\displaystyle P(A\mid E_2)=\dfrac{10}{100}$ or $\displaystyle \dfrac{1}{10}$
(ii)
$\displaystyle P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)$
$\displaystyle =\dfrac{40}{100}\times\dfrac{80}{100}+\dfrac{60}{100}\times\dfrac{10}{100}$
$\displaystyle =\dfrac{38}{100}$ or $\displaystyle \dfrac{19}{50}$
(iii)
(a)
$\displaystyle P(E_1\mid A)=\dfrac{P(E_1)P(A\mid E_1)}{P(A)}$
$\displaystyle =\dfrac{\dfrac{40}{100}\times\dfrac{80}{100}}{\dfrac{38}{100}}=\dfrac{16}{19}$
OR
(b)
$\displaystyle P(E_2\mid A)=\dfrac{P(E_2)P(A\mid E_2)}{P(A)}$
$\displaystyle =\dfrac{\dfrac{60}{100}\times\dfrac{10}{100}}{\dfrac{38}{100}}=\dfrac{3}{19}$
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.