Conditional Probability

Question

A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices.

As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer?

Comments

I am getting 0.787 I.e. 78.7 % but answer is 77.4 %. Can someone verify 

 

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https://math.stackexchange.com/questions/2348963/conditional-probability-regarding-multiple-choice

Answer 1

This problem will be solved using Bayes Theorem.

Let C be the event that the student answered corectly

P(C) = $\frac{1}{2}+\frac{1}{4}*\frac{1}{3}+\frac{1}{4}*\frac{1}{4} = \frac{31}{48}$

Let X be the event that student knew the answer.

Then , P(X/C) = $\frac{P(X)*P(C/X)}{P(C)} = \frac{\frac{1}{2}*1}{\frac{31}{48}} = \frac{48}{62} = 0.7741$.

Thus in percentage = 77.41%

Comments

Why did u take 1/4 again in second case...isn't it 3/4

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1/4+3/4=1 .. is this correct? Should be 1/2 right?

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Got it... Thank You

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Brother can u point out the mistake in my solution

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I mean when the student doesn't know the answer then he'll eliminate the option right so shouldn't it be 0.5*0.25*0.33   for getting correct option after eliminating one option.

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0.5+0.25+0.25 = 1 , right? Why are you multiplying 0.5 again ? 0.5 is getting divided into two sub cases right?

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according TO ABOVE TREE DIAGRAM:

SEE What i did:

By probability 1/2 he didn't knew the answer

When he didn't knew the answer by probability 0.25 he will be able to eliminate one option

and after eliminating one option probability of choosing correct option is 1/3

Hence = 0.5 * 0.25 * 0.33 ----> When he didn't knew the answer and then he eliminates one option and chooses correct option.

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But out of 50% of the time , 25% of the times he can eliminate one option and 25% of the times he cannot.

Multiplying 50% in both the cases , is this correct , think about it?

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I thought like 25% of 50% he can eliminate the option and 75% of 50% he can't eliminate. That's why I multiplied by 0.5.

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25% of the time he can eleminate and 25% of the time he cannot , sum of both the possibilities should be 50% right?

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i feel language isn't clear in the first go..

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what did you not understand?

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its clear now .

that time didnt read the question properly