A multiple choice examination has 5 questions. Each question has three alternative answers of which

Q: Wait, why do we use the Binomial Distribution here?

Great question! We use the Binomial Distribution because the problem satisfies three key criteria: the number of trials is fixed ($n=5$), each trial is independent (guessing one question doesn't change the odds for the next), and each trial has only two possible outcomes (success or failure).

Q: Why is the probability of failure $q = 2/3$?

In any probability experiment, the sum of all possible outcomes must equal $1$. Since there are $3$ alternatives and only $1$ is correct, the probability of success $p$ is $1/3$. Therefore, the probability of failure $q$ must be $1 - p$, which is $1 - 1/3 = 2/3$.

Q: Why do we add $P(X=4)$ and $P(X=5)$?

Because these two events—getting exactly $4$ correct and getting exactly $5$ correct—are mutually exclusive. You cannot get exactly $4$ correct and exactly $5$ correct at the same time. When events are mutually exclusive, the probability of either occurring is the sum of their individual probabilities.

Question

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

Options:

(A) $17/3^5$

(B) $13/3^5$

(C) $11/3^5$

(D) $10/3^5$

Sigma Percentile · Accepted Answer

### The Gambler's Dilemma: A Mathematical Reality Check

Imagine you are sitting in a high-stakes examination hall. You have five questions in front of you, and for each one, you have absolutely no idea what the answer is.

You decide to rely purely on luck, guessing randomly for every single question. It feels like a simple game of chance, but beneath the surface lies a rigid, beautiful mathematical structure. Today, we are going to peel back the curtain on this scenario using the power of the **Binomial Distribution**.

### Defining the Battlefield

First, let us define our parameters. We have $n = 5$ questions. Each question is an independent trial, meaning whether you guess correctly on the first question has zero impact on the second.

For each question, there are three alternatives, and only one is correct. This means the probability of success, which we denote as $p$, is $\frac{1}{3}$.

Consequently, the probability of failure, $q$, is the complement:
$$q = 1 - p = \frac{2}{3}$$

### The Binomial Engine

To find the probability of any specific number of successes, we use the Binomial Distribution formula:

$$P(X = k) = \binom{n}{k} p^k q^{n-k}$$

Here, $\binom{n}{k}$ (or $nCk$) tells us the number of ways to choose which $k$ questions out of $n$ are answered correctly. It is the combinatorial heart of the formula.

Our target is to find the probability of getting $4$ or more correct answers, which corresponds to the event $X \ge 4$. Since our outcomes are discrete, this is simply the sum of two distinct possibilities: getting exactly $4$ correct or getting exactly $5$ correct.

$$P(X \ge 4) = P(X = 4) + P(X = 5)$$

### The Calculation

Let us tackle these one by one. For $k = 4$, we substitute our values into the formula:

$$P(X = 4) = \binom{5}{4} \left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^1$$

We know that $\binom{5}{4} = 5$. So, the calculation becomes:

$$P(X = 4) = 5 \cdot \frac{1}{3^4} \cdot \frac{2}{3} = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{3^5}$$

Now, for the rare event of getting all $5$ correct ($k = 5$):

$$P(X = 5) = \binom{5}{5} \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^0$$

Since $\binom{5}{5} = 1$ and any number to the power of $0$ is $1$, this simplifies beautifully:

$$P(X = 5) = 1 \cdot \frac{1}{3^5} \cdot 1 = \frac{1}{3^5}$$

### The Final Synthesis

Now, we bring them together. The total probability is the sum of these two results:

$$P(X \ge 4) = \frac{10}{3^5} + \frac{1}{3^5} = \frac{11}{3^5}$$

This fraction, **$\frac{11}{243}$**, is our final answer. If you calculate the value, it is approximately $0.045$, or about **$4.5\%$**.

Think about that for a moment. If you walk into an exam and guess on every question, you have less than a $5\%$ chance of getting $4$ or more correct. Mathematics has just quantified the risk of guessing—it is a strategy that is almost certainly doomed to fail. Let this be a reminder that in the JEE, as in life, preparation beats probability every single time.

Animated Solution for Mathematics - Probability: A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

Select Answer:

Visualized Solution

Visualizing the Setup

Success and Failure Probabilities

The Binomial Formula

Target Event: $P (X \geq 4)$

Calculating $P (X = 4)$

Calculating $P (X = 5)$

Combining the Probabilities

Final Conclusion

The Sigma Insight: Binomial Distribution

The Gambler's Dilemma

Defining the Battlefield

The Binomial Engine

The Calculation

The Final Synthesis

Similar Questions

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$ , so that the probability of guessing at least ' $n$ ' correct answers is less than $1/2$ , is

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to:

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 4/5, then the probability that he is unable to solve less than two problems is :

The probability that an event $A$ happens in one trial of an experiment is $0.4$ . Three independent trials of the experiment are performed. The probability that the event $A$ happens at least once is

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $\frac{k}{3 ^{11}}$ , then $k$ is equal to

Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$ . Then $54 P (X \leq 2)$ is equal to

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is

The mean and variance of a binomial distribution are $α$ and $\frac{α}{3}$ respectively. If $P (X = 1) = \frac{4}{243}$ , then $P (X = 4 or 5)$ is equal to:

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is