In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer

Q: Wait, why can't we just use one binomial distribution formula for all 10 questions?

That is a classic trap! You can only use a single binomial distribution if the probability of success is uniform across all trials. Here, the probability of success changes after the 4th question. You have two distinct groups with different success rates, so you must treat them as independent binomial variables and combine them.

Q: How do I know I haven't missed a case?

Always check your boundaries. $X$ cannot exceed 4, and $Y$ cannot exceed 6. If you try $X=1$, then $Y=7$, which is impossible because there are only 6 questions in Group B. The physical constraints of the problem define the limits of your search, so once you hit those boundaries, you know you are done.

Q: Is there a shortcut to calculating the final sum?

While advanced techniques like generating functions exist, for a small number of cases like this, direct calculation is safer and less prone to algebraic errors. The systematic case-by-case approach is the most reliable way to ensure you don't make a silly calculation error in the exam hall.

Question

In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability $\frac{3}{4}$ and the remaining 6 questions correctly with probability $\frac{1}{4}$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is $\frac{27k}{4^{10}}$, then $k$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are sitting in the examination hall with $10$ questions. For the first $4$ questions, your success probability is $p_1 = \frac{3}{4}$. For the remaining $6$ questions, your success probability is $p_2 = \frac{1}{4}$.

We define $X$ as the number of correct answers from the first group (Group A) and $Y$ as the number of correct answers from the second group (Group B). We seek the probability of getting exactly $8$ questions correct, which implies the constraint $X + Y = 8$.

### The Constraint Logic

Given the constraints $0 \le X \le 4$ and $0 \le Y \le 6$, we identify the valid integer partitions for $X + Y = 8$:

1. Case 1: $X=4$ and $Y=4$
2. Case 2: $X=3$ and $Y=5$
3. Case 3: $X=2$ and $Y=6$

These are the only three mutually exclusive paths to achieving exactly $8$ correct answers.

### The Binomial Engine

We utilize the Binomial Distribution formula: $P(X=r) = \binom{n}{r} p^r q^{n-r}$. We calculate the probability for each case:

For Case 1 ($X=4, Y=4$):
$$P_1 = \binom{4}{4} \left(\frac{3}{4}\right)^4 \cdot \binom{6}{4} \left(\frac{1}{4}\right)^4 \left(\frac{3}{4}\right)^2 = 1 \cdot \frac{3^4}{4^4} \cdot 15 \cdot \frac{3^2}{4^6} = \frac{15 \cdot 3^6}{4^{10}}$$

For Case 2 ($X=3, Y=5$):
$$P_2 = \binom{4}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^1 \cdot \binom{6}{5} \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^1 = 4 \cdot \frac{3^3}{4^4} \cdot 6 \cdot \frac{3^1}{4^6} = \frac{24 \cdot 3^4}{4^{10}}$$

For Case 3 ($X=2, Y=6$):
$$P_3 = \binom{4}{2} \left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^2 \cdot \binom{6}{6} \left(\frac{1}{4}\right)^6 = 6 \cdot \frac{3^2}{4^4} \cdot 1 \cdot \frac{1}{4^6} = \frac{6 \cdot 3^2}{4^{10}}$$

### The Grand Summation

The total probability is the sum of these cases:
$$P_{total} = \frac{15 \cdot 3^6 + 24 \cdot 3^4 + 6 \cdot 3^2}{4^{10}}$$

Factoring out $3^2 = 9$ from the numerator:
$$P_{total} = \frac{9(15 \cdot 3^4 + 24 \cdot 3^2 + 6)}{4^{10}} = \frac{9(15 \cdot 81 + 24 \cdot 9 + 6)}{4^{10}} = \frac{9(1215 + 216 + 6)}{4^{10}} = \frac{9(1437)}{4^{10}} = \frac{12933}{4^{10}}$$

We are given that the probability is $\frac{27k}{4^{10}}$. Equating the numerators:
$$27k = 12933$$

Dividing by $27$:
$$k = \frac{12933}{27} = 479$$

The final value is **$k = 479$**.

Enter Numerical Value:

Visualized Solution

Defining the Two Groups

The Objective: Exactly $8$ Correct

Identifying Possible Cases

Case 1: $4$ from A and $4$ from B

Case 2: $3$ from A and $5$ from B

Case 3: $2$ from A and $6$ from B

Summing the Probabilities

Simplifying the Numerator

Calculating the Final Sum

Solving for $k$

Final Answer: $k = 479$

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Constraint Logic

The Binomial Engine

The Grand Summation

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