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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

38. In answering a question on a multiple choice test a student either knows the answer or guesses. Let 3/4 be the probability that he knows the answer and 1/4 be the probability that he guesses. Assuming that a student who guesses the answer, will be correct with probability 1/4. What is the probability that a student knows the answer given that he answered it correctly?

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alok kumar singh | Contributor-Level 10

4 months ago

38. Let, $E_{1}$ :event that knows the answer

$E_{2}$ :event that he guesses the answer

$P (E_{1}) = \frac{3}{4}$ and $P (E_{2}) = \frac{1}{4}$

Let $A$ be the event that the answer is correct

Also, $P (A / E_{1}) = P$ (correct answer given that he knows) $= 1$

$P (A / E_{2}) = P$ (correct answer given that he guess) $= \frac{1}{4}$

Now, probability that he knows the answer given that he answer it correctly is $P (E_{1} / A)$ ,

By Baye’s theorem,

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2})}$

$= \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}}$

$= \frac{\frac{3}{4}}{\frac{12 + 1}{16}} = \frac{\frac{3}{4}}{\frac{13}{16}}$

$= \frac{3}{4} \times \frac{16}{13} = \frac{12}{13}$

$P (E_{1} / A) = \frac{12}{13}$

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ax² + bx + c = 0

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