For an initial screening of an admission test, a candidate is given fifty problems to solve. If the

Q: Wait, why do we define $X$ as 'unable to solve'? Can't we just use 'solve'?

You absolutely could! But if you define $X$ as 'number of problems solved', the condition 'unable to solve less than two' becomes 'number of problems solved is greater than 48'. That would mean calculating $P(X=49) + P(X=50)$, which is also correct but slightly more tedious. Defining $X$ as 'unable to solve' makes the condition $X < 2$ (i.e., $X=0$ or $X=1$), which is much simpler to calculate.

Q: Why is $X < 2$ only $0$ and $1$? Why not $2$?

Great catch! The phrase 'less than two' means strictly less than two. In mathematics, this is denoted as $X < 2$. Since $X$ represents the number of problems, it must be a non-negative integer. The integers strictly less than 2 are 0 and 1. If the question had said 'at most two', then we would have included 2.

Q: How do I know when to use the Binomial Distribution?

Look for these three signs: 1) A fixed number of trials ($n=50$). 2) Each trial has only two possible outcomes (success/failure). 3) The probability of success is constant for each trial. When you see these, the Binomial Distribution is your best friend!

Question

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 :

Options:

(A) $(316/25)(4/5)^{48}$

(B) $(54/5)(4/5)^{49}$

(C) $(164/25)(1/5)^{48}$

(D) $(201/5)(1/5)^{49}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a classic probability challenge that often trips up even the most prepared students. Imagine you are standing in an exam hall, staring at fifty problems.

The problem states that for any given problem, the probability of solving it is $P(	ext{solve}) = \frac{4}{5}$. However, the question asks for the probability that the candidate is "unable to solve less than two problems."

This phrasing is a test of your ability to reframe the problem. Instead of focusing on what the candidate *can* do, let us focus on what they *cannot* do.

Let $X$ be the random variable representing the number of problems the candidate is **unable to solve**. If the probability of solving a problem is $\frac{4}{5}$, then the probability of failing to solve it is $p = 1 - \frac{4}{5} = \frac{1}{5}$.

Conversely, the probability of "success" in our defined variable $X$ (which is failing to solve) is $p = \frac{1}{5}$, and the probability of "failure" (solving the problem) is $q = \frac{4}{5}$. This simple redefinition is the key to unlocking the entire problem.

### The Binomial Toolkit

Now that we have our parameters, $n = 50$, $p = \frac{1}{5}$, and $q = \frac{4}{5}$, we need to interpret the condition. The question asks for the probability that the candidate is unable to solve **less than two** problems.

Mathematically, this is $X < 2$. Since $X$ is a discrete count of problems, it can only take non-negative integer values. Therefore, $X < 2$ implies $X = 0$ or $X = 1$.

We are looking for the sum of these two probabilities:
$$P(X < 2) = P(X = 0) + P(X = 1)$$

To calculate these, we use the Binomial Distribution formula:
$$P(X = r) = \binom{n}{r} p^r q^{n-r}$$

This formula is the backbone of discrete probability. It tells us the likelihood of getting exactly $r$ successes in $n$ trials.

### The Calculation

First, for $X = 0$, the candidate is unable to solve zero problems, meaning they solved all of them. Plugging $r = 0$ into our formula:
$$P(X = 0) = \binom{50}{0} \left(\frac{1}{5}ight)^0 \left(\frac{4}{5}ight)^{50}$$

Since $\binom{50}{0} = 1$ and any number to the power of zero is 1, this simplifies beautifully to:
$$P(X = 0) = \left(\frac{4}{5}ight)^{50}$$

Next, for $X = 1$, the candidate is unable to solve exactly one problem. Plugging $r = 1$ into our formula:
$$P(X = 1) = \binom{50}{1} \left(\frac{1}{5}ight)^1 \left(\frac{4}{5}ight)^{49}$$

We know that $\binom{50}{1} = 50$. So, $P(X = 1) = 50 \cdot \frac{1}{5} \cdot \left(\frac{4}{5}ight)^{49}$. Simplifying $50 \cdot \frac{1}{5}$ gives us 10. Thus:
$$P(X = 1) = 10 \cdot \left(\frac{4}{5}ight)^{49}$$

### Final Calculation

Now, we combine these two results:
$$P(X < 2) = \left(\frac{4}{5}ight)^{50} + 10 \cdot \left(\frac{4}{5}ight)^{49}$$

To reach the final answer, we factor out the common term, $\left(\frac{4}{5}ight)^{49}$:
$$P(X < 2) = \left(\frac{4}{5}ight)^{49} \left[ \frac{4}{5} + 10 ight]$$

Inside the bracket, we have $\frac{4}{5} + 10 = \frac{4 + 50}{5} = \frac{54}{5}$. Substituting this back, we get the final probability:
$$P(X < 2) = \frac{54}{5} \left(\frac{4}{5}ight)^{49}$$

The beauty of this problem lies in the redefinition of the random variable. By shifting our perspective from "solving" to "unable to solve," we turned a potentially complex calculation into a simple, elegant binomial sum. Remember, in JEE Advanced, the most powerful tool you have is your ability to frame the problem in the most efficient way possible.

Animated Solution for Mathematics - Probability: 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 :

Select Answer:

Visualized Solution

Problem Parameters

Defining the Random Variable

Probability of Failure

Interpreting the Condition

Possible Values of $X$

Binomial Probability Formula

Setting up $P (X = 0)$

Computing $P (X = 0)$

Setting up $P (X = 1)$

Computing $P (X = 1)$

Combining the Probabilities

Factorizing the Expression

Final Arithmetic

Conclusion

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Binomial Toolkit

The Calculation

Final Calculation

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