A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guessed it, is 1/2 Also assume that the probability of the answer for a question being guessed, given that the student’s answer is correct, is 1/6 Then the probability that the student knows the answer of a randomly chosen question is

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 Alright, let's tackle this probability problem step by step. I'll try to break it down and understand what's given and what's being asked.

Understanding the Problem

We have a student taking a true-false quiz. For each question:

  1. Knows the answer: The student definitely answers correctly.

  2. Guesses the answer: The student has a 1/2 chance of getting it right (since it's true-false).

We're given:

  • P(correct | guessed) = 1/2

  • P(guessed | correct) = 1/6

We need to find:

  • P(knows the answer) for a randomly chosen question.

Let's define some events to make it clearer:

  • Let K be the event that the student knows the answer.

  • Let G be the event that the student guesses the answer. (Note: G = not K, since the student either knows or guesses.)

  • Let C be the event that the answer is correct.

Given:

  1. P(C | G) = 1/2

  2. P(G | C) = 1/6

We need to find P(K).

Using Conditional Probability and Bayes' Theorem

First, recall that P(G | C) = P(C | G) * P(G) / P(C). This is Bayes' Theorem.

We know P(C | G) = 1/2 and P(G | C) = 1/6.

So:
1/6 = (1/2) * P(G) / P(C)

Thus:
P(C) = (1/2) * P(G) / (1/6) = (1/2) * P(G) * 6 = 3 * P(G)

Now, how is P(C) defined overall? The student can be correct either by knowing the answer or by guessing correctly.

So:
P(C) = P(C | K) * P(K) + P(C | G) * P(G)

We know:

  • If the student knows the answer, they are definitely correct: P(C | K) = 1

  • P(C | G) = 1/2

So:
P(C) = 1 * P(K) + (1/2) * P(G)

But P(G) = 1 - P(K), since the student either knows or guesses.

So:
P(C) = P(K) + (1/2)(1 - P(K)) = P(K) + 1/2 - (1/2)P(K) = (1/2)P(K) + 1/2

Earlier, we had P(C) = 3 * P(G) = 3 * (1 - P(K))

So:
(1/2)P(K) + 1/2 = 3(1 - P(K))

Now, let's solve for P(K):

Multiply both sides by 2 to eliminate fractions:
P(K) + 1 = 6(1 - P(K))

Expand the right side:
P(K) + 1 = 6 - 6 P(K)

Now, gather like terms:
P(K) + 6 P(K) = 6 - 1
7 P(K) = 5

So:
P(K) = 5/7

Verifying the Answer

Let's check if this makes sense.

Given P(K) = 5/7, then P(G) = 2/7.

Compute P(C):
P(C) = P(K) + (1/2)P(G) = 5/7 + (1/2)(2/7) = 5/7 + 1/7 = 6/7

Now, compute P(G | C) using Bayes':
P(G | C) = P(C | G) * P(G) / P(C) = (1/2)*(2/7) / (6/7) = (1/7) / (6/7) = 1/6

This matches the given P(G | C) = 1/6, so our answer is consistent.

Final Answer

The probability that the student knows the answer of a randomly chosen question is 5/7.

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