In a multiple-choice question there are four alternative answers, of which one or more are correct.

Q: Wait, why is the total number of ways 15 and not 16? Why do we subtract 1?

That is a fantastic question! In a multiple-choice question where 'one or more' options can be correct, each option acts like a binary switch: it is either 'ticked' or 'unticked'. With 4 options, that gives us $2^4 = 16$ possible ways to mark the paper. However, the problem implies that the student is submitting an answer. If the student ticks nothing, they haven't actually answered the question. Therefore, we must exclude the 'empty set' case where all options are left blank. That leaves us with $16 - 1 = 15$ valid ways to answer.

Q: Why can't we just use $P = \frac{3}{16}$? Why is the denominator 15?

I understand the confusion! It is tempting to use the total number of combinations (16), but remember, the student is choosing from the set of *valid* answers. Since the student cannot submit a blank answer, the sample space of 'possible answers' is restricted to the 15 valid combinations. If you include the blank option, you are counting an impossible event as a potential choice, which would skew your probability. Always define your sample space based on the constraints of the problem!

Q: Is the intuitive method (3/15) always valid? It feels too simple.

It feels like magic, doesn't it? The intuitive method works here because the selection is random and without replacement. Think of it this way: you have 15 slips of paper in a hat, and only 1 is the 'winning' slip. You are allowed to pull out 3 slips. The probability that the winning slip is among the 3 you pulled is simply the number of your draws divided by the total number of slips. It is a beautiful shortcut, but always ensure the 'winning' condition is unique before applying it!

Question

In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidate decides to tick the answers at random, if he is allowed upto three chances to answer the questions, find the probability that he will get marks in the questions.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are sitting in the examination hall for the JEE Advanced. You encounter a multiple-choice question where "one or more" answers could be correct, with four options: $A$, $B$, $C$, and $D$. You are allowed three chances to guess.

### Defining the Universe of Possibilities

First, we must define our sample space. If each option can either be ticked or left blank, we have two states for each of the four options. Mathematically, this is:

$$2 	imes 2 	imes 2 	imes 2 = 2^4 = 16$$

However, the problem implies a valid submission. If you tick nothing, you have not answered the question, so we must discard the case where all options are blank. Our total valid sample space is $16 - 1 = 15$.

### The Strategy of the Complement

We need to find the probability of getting marks within three chances. Calculating the probability of success on the first, second, or third try individually is tedious and prone to error.

Instead, we use the **Complement Rule**:

$$P(	ext{Success}) = 1 - P(	ext{Failure})$$

It is much easier to calculate the probability that you fail all three times and subtract that from certainty.

### The Math of Failure

To fail all three times, you must pick three combinations from the 15 available, and all three must be incorrect. Since there is only $1$ correct combination, there are $15 - 1 = 14$ incorrect combinations.

The number of ways to choose 3 incorrect combinations from 14 is given by the combination formula ${}^{14}C_3$. The total number of ways to choose any 3 combinations from the 15 available is ${}^{15}C_3$. Thus, the probability of failing all three times is:

$$P(	ext{Failure}) = \frac{{}^{14}C_3}{{}^{15}C_3}$$

Expanding this using factorials, we observe the following:

$$P(	ext{Failure}) = \frac{\frac{14 	imes 13 	imes 12}{3!}}{\frac{15 	imes 14 	imes 13}{3!}}$$

The $3!$ in the denominator of both the numerator and the denominator cancels out. Furthermore, the terms $14$ and $13$ appear in both, so they vanish as well. We are left with:

$$P(	ext{Failure}) = \frac{12}{15}$$

Simplifying this by dividing both the numerator and denominator by $3$, we get:

$$P(	ext{Failure}) = \frac{4}{5}$$

### The Final Victory

Now, we apply our complement rule to find the probability of success:

$$P(	ext{Success}) = 1 - P(	ext{Failure}) = 1 - \frac{4}{5} = \frac{1}{5}$$

The probability of getting marks is **$\frac{1}{5}$**.

### The Intuitive Shortcut

There is a faster way to visualize this. Imagine 15 slips of paper in a bowl, where only 1 is the winning ticket.

If you are allowed to draw 3 slips, the probability that the winning slip is in your hand is simply the number of slips you draw divided by the total number of slips:

$$\frac{3}{15} = \frac{1}{5}$$

Both the rigorous combinatorial method and this intuitive approach lead us to the same truth. In the JEE, understanding the "why" behind the math is what separates the good from the great.

Visualized Solution

Visualizing the Options

Total Ticking Combinations

Correct vs Incorrect Combinations

The Candidate's Chances

Defining the Strategy

Setting up the Failure Probability

Calculating the Failure Probability

Simplifying the Fraction

The Final Result

Alternative Intuitive Method

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Defining the Universe of Possibilities

The Strategy of the Complement

The Math of Failure

The Final Victory

The Intuitive Shortcut

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