A student appears for a quiz consisting of only true-false type questions and answers all the questi

Q: Wait, why is $P(A|E_1) = 1$? Doesn't the student ever make a mistake?

Great question! In the context of this problem, 'knowing' implies absolute certainty. If the student knows the answer, they are 100% correct. We are modeling the ideal state of knowledge versus the random state of guessing. Don't overcomplicate the model—stick to the definitions given in the problem statement!

Q: Why can't we just use $P(E_2) = \frac{1}{6}$ directly?

That is a common trap! $\frac{1}{6}$ is the probability of guessing *given* that the answer is correct. It is a conditional probability, not the prior probability of guessing. You must use Bayes' Theorem to 'reverse' the condition and find the prior probability $x$.

Q: Why is the denominator $1 - \frac{x}{2}$?

Let's look at the total probability of being correct: $P(A) = (1-x)(1) + (x)(\frac{1}{2})$. Expanding this gives $1 - x + \frac{x}{2}$. Combining the $x$ terms, $-x + \frac{x}{2}$ becomes $-\frac{x}{2}$. Thus, we get $1 - \frac{x}{2}$. It is just simple algebraic simplification!

Question

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 guess it, is $\frac{1}{2}$. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is $\frac{1}{6}$. Then the probability that the student knows the answer of a randomly chosen question is

Options:

(A) $\frac{1}{12}$

(B) $\frac{1}{7}$

(C) $\frac{5}{7}$

(D) $\frac{5}{12}$

Sigma Percentile · Accepted Answer

### The Detective's Dilemma: Unmasking the Truth

Imagine you are a detective investigating a quiz. You have a student who has just submitted their answers. You see a correct answer on the paper, but you are faced with a mystery: Did the student actually know the answer, or did they just get lucky with a guess?

This is not just a math problem; it is a fundamental question of inference. We are looking at an effect—a correct answer—and trying to trace it back to its cause. This is the heart of Bayesian probability.

### Phase 1: The Tree of Possibilities

To solve this, we must first map out the student's mental state. We define two mutually exclusive events. Let $E_1$ be the event that the student knows the answer, and $E_2$ be the event that the student guesses.

Since the student must either know or guess, these two events partition the sample space. If we let the probability of guessing be $P(E_2) = x$, then the probability of knowing must be $P(E_1) = 1 - x$. Think of this as a fork in the road; every question starts here.

### Phase 2: The Conditional Reality

Now, we introduce the outcome: event $A$, which is the student getting the answer correct. We need to understand how $A$ relates to our two paths.

If the student knows the answer ($E_1$), they are infallible. Thus, the probability of being correct given they know is $P(A|E_1) = 1$.

However, if they guess ($E_2$), they are at the mercy of chance. Since it is a true-false question, they have a $50\%$ chance of being right. So, $P(A|E_2) = \frac{1}{2}$. This is our conditional framework.

### Phase 3: The Bayesian Shift

Here is where the magic happens. The problem gives us a piece of information that feels backwards: 'The probability of the answer being guessed, given that the student's answer is correct, is $\frac{1}{6}$.' Mathematically, this is $P(E_2|A) = \frac{1}{6}$.

We are looking at the result and asking about the cause. This is the exact scenario for Bayes' Theorem. The formula is elegant and powerful:

$$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$$

This formula is essentially a ratio of the 'guessing path' to the 'total probability of being correct.' It allows us to isolate the probability of guessing from the noise of the total correct answers.

### Phase 4: The Algebraic Dance

Now, let us substitute our values into this equation. On the left, we have $\frac{1}{6}$. On the right, the numerator is the guessing path: $x \cdot \frac{1}{2}$. The denominator is the sum of both paths: $(1-x) \cdot 1 + x \cdot \frac{1}{2}$.

Putting it all together, we get:

$$\frac{1}{6} = \frac{x \cdot \frac{1}{2}}{(1-x) + \frac{x}{2}}$$

Let us simplify the denominator. We have $1 - x + \frac{x}{2}$, which simplifies to $1 - \frac{x}{2}$. Now our equation looks like this:

$$\frac{1}{6} = \frac{\frac{x}{2}}{1 - \frac{x}{2}}$$

To clear the fractions, we multiply the numerator and denominator on the right by $2$. This transforms the equation into:

$$\frac{1}{6} = \frac{x}{2 - x}$$

Now, we cross-multiply. $1 \cdot (2 - x) = 6 \cdot x$, which simplifies to $2 - x = 6x$. Adding $x$ to both sides, we find $7x = 2$, or $x = \frac{2}{7}$.

### The Final Revelation

We have found $x$, which is the probability that the student guessed. But do not stop here! The question asks for the probability that the student **knows** the answer.

That is $P(E_1) = 1 - x$. Substituting our value, we get:

$$1 - \frac{2}{7} = \frac{5}{7}$$

And there it is. The probability that the student knew the answer is **$\frac{5}{7}$**. We started with uncertainty, applied the rigorous logic of Bayes, and arrived at a precise, deterministic answer.

Select Answer:

Visualized Solution

Defining the Initial Events

Conditional Probabilities of Correctness

The Reverse Probability (Bayes' Theorem)

Substituting the Values

Simplifying the Denominator

Clearing the Fractions

Cross Multiplication

Solving for $x$

Finding the Final Probability

The Sigma Insight: Bayes' Theorem

The Detective's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

The Final Revelation

Similar Questions

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Comprehension Passage

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :