CBSE 2025, Set 3

Question

**Case Study - 1**
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 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, 5 students participated in a javelin throw competition. The distances to which they have thrown the javelin are shown below in the table:

| Name of student | Distance of javelin (in meters) |
|---|---|
| Ajay | $47.7$ |
| Bijoy | $47.07$ |
| Kartik | $43.09$ |
| Dinesh | $43.9$ |
| Devesh | $45.2$ |

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 $40\%$ of the students have the misconception in the concept of decimal comparison and the rest do not have the misconception. $80\%$ of the students having misconception answered Bijoy as the correct answer in the paper. $90\%$ 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 question:
(iii) (a) What is the probability that a student who answered as Bijoy is having misconception?

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Define the required conditional probability. Let $E_1$ = Event that the student has the misconception. Let $E_2$ = Event that the student does not have the misconception. Let $A$ = Event that the student answers 'Bijoy'. We need to find the probability that the student has the misconception GIVEN that they answered Bijoy. This is represented by $P(E_1|A)$. **Step 2:** Apply Bayes' Theorem. $$P(E_1|A) = \frac{P(E_1 \cap A)}{P(A)} = \frac{P(E_1) 	imes P(A|E_1)}{P(E_1) 	imes P(A|E_1) + P(E_2) 	imes P(A|E_2)}$$ **Step 3:** Substitute the previously calculated values. From the previous parts, we know: $P(E_1) 	imes P(A|E_1) = 0.40 	imes 0.80 = 0.32$ (Numerator) $P(A) = 0.38$ (Denominator - total probability of answering Bijoy) $$P(E_1|A) = \frac{0.32}{0.38}$$ **Step 4:** Simplify the fraction. $$P(E_1|A) = \frac{32}{38} = \frac{16}{19}$$ Therefore, the probability is **$\frac{16}{19}$**.

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