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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 5 · Set 1 · Q35
(a) In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 3/5 be the probability that he knows the…
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\displaystyle \frac{3}{5}$ be the probability that he knows the answer and $\displaystyle \frac{2}{5}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\displaystyle \frac{1}{3}$. What is the probability that the student knows the answer, given that he answered it correctly?
A box contains $\displaystyle 10$ tickets, $\displaystyle 2$ of which carry a prize of ₹ $\displaystyle 8$ each, $\displaystyle 5$ of which carry a prize of ₹ $\displaystyle 4$ each, and remaining $\displaystyle 3$ carry a prize of ₹ $\displaystyle 2$ each. If one ticket is drawn at random, find the mean value of the prize.
Marking-scheme solution
Let events $\displaystyle \mathrm{A}, \mathrm{B}$ and E be defined as:
A : Student knows the answer
B : Student guesses the answer
E : student answered correctly
$$\mathrm{P}(\mathrm{~A})=\frac{3}{5}, \mathrm{P}(\mathrm{~B})=\frac{2}{5}
$$
Here, $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{A}}\right)=1$ and $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}}{\mathrm{B}}\right)=\frac{1}{3}$
By Bayes' Theorem
$$\mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{E}}\right)=\frac{\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~A}}\right)}{\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~A}}\right)+\mathrm{P}(\mathrm{~B}) \cdot \mathrm{P}\left(\dfrac{\mathrm{E}}{\mathrm{~B}}\right)}
$$
Let X denote the prize value.
Here X can take values of $\displaystyle 8,4$ and 2.
$$\mathrm{P}(\mathrm{X}=8)=\frac{2}{10}, \text { or } \frac{1}{5}
$$
$$\mathrm{P}(\mathrm{X}=4)=\frac{5}{10}, \text { or } \frac{1}{2}
$$
$$\mathrm{P}(\mathrm{X}=2)=\frac{3}{10}
$$
Hence, Mean value of $\displaystyle \mathrm{X}=\sum \mathrm{X} \mathrm{P}(\mathrm{X})=\frac{8}{5}+2+\frac{6}{10}$
| X | $\displaystyle 8$ | $\displaystyle 4$ | $\displaystyle 2$ |
| $\displaystyle \mathrm{P}(\mathrm{X})$ | $\displaystyle \frac{1}{5}$ | $\displaystyle \frac{1}{2}$ | $\displaystyle \frac{3}{10}$ |
| $\displaystyle \mathrm{XP}(\mathrm{X})$ | $\displaystyle \frac{8}{5}$ | $\displaystyle \frac{4}{2}$ | $\displaystyle \frac{6}{10}$ |
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.