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Step-by-Step Solution
Step 1: Understand the Problem Setup
We have 10 true/false questions. Among these:
4 questions (call this set A) each have a probability of being guessed correctly as \frac{3}{4} .
6 questions (call this set B) each have a probability of being guessed correctly as \frac{1}{4} .
We want the probability that the student guesses exactly 8 questions correctly. It is given that this probability equals
\displaystyle \frac{27\,k}{4^{10}}. Our goal is to find the value of k .
Step 2: Identify Possible Ways to Get 8 Correct
The total of 8 correct answers can be distributed between the two sets (A and B) in the following ways:
All 4 are correct in set A, and exactly 4 are correct in set B.
3 are correct in set A, and exactly 5 are correct in set B.
2 are correct in set A, and exactly 6 are correct in set B.
We will calculate the probability of each case and then sum them up.
Step 3: Probability of Each Case
Case (i): 4 Correct from Set A and 4 Correct from Set B
Number of ways to choose 4 correct out of the 4 in set A: C(4,4) = 1 .
Probability all 4 in set A are correct: \left(\frac{3}{4}\right)^4 .
Number of ways to choose 4 correct out of the 6 in set B: C(6,4) .
Probability these 4 in set B are correct: \left(\frac{1}{4}\right)^4 .
Probability the remaining 2 in set B are incorrect: \left(\frac{3}{4}\right)^2 .
So, the probability of Case (i) is:
C(4,4)\,\left(\frac{3}{4}\right)^4 \times
C(6,4)\,\left(\frac{1}{4}\right)^4 \left(\frac{3}{4}\right)^2.
Case (ii): 3 Correct from Set A and 5 Correct from Set B
Number of ways to choose 3 correct out of 4 in set A: C(4,3) .
Probability those 3 are correct: \left(\frac{3}{4}\right)^3 .
Probability the remaining 1 in set A is incorrect: \left(\frac{1}{4}\right)^1 .
Number of ways to choose 5 correct out of 6 in set B: C(6,5) .
Probability those 5 are correct: \left(\frac{1}{4}\right)^5 .
Probability the remaining 1 in set B is incorrect: \left(\frac{3}{4}\right)^1 .
So, the probability of Case (ii) is:
C(4,3)\,\left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^1
\times
C(6,5)\,\left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^1.
Case (iii): 2 Correct from Set A and 6 Correct from Set B
Number of ways to choose 2 correct out of 4 in set A: C(4,2) .
Probability those 2 are correct: \left(\frac{3}{4}\right)^2 .
Probability the remaining 2 in set A are incorrect: \left(\frac{1}{4}\right)^2 .
Number of ways to choose 6 correct out of 6 in set B: C(6,6) = 1 .
Probability those 6 are correct: \left(\frac{1}{4}\right)^6 .
So, the probability of Case (iii) is:
C(4,2)\,\left(\frac{3}{4}\right)^2 \left(\frac{1}{4}\right)^2
\times
C(6,6)\,\left(\frac{1}{4}\right)^6.
Step 4: Sum of the Probabilities
The total probability that exactly 8 questions are guessed correctly is:
P(\text{exactly 8 correct})
= \text{(Case (i))} + \text{(Case (ii))} + \text{(Case (iii))}.
After simplification, this sum matches the given form
\displaystyle \frac{27\,k}{4^{10}}.
Step 5: Find the Value of k
Comparing both sides and expanding the combinations and probabilities leads to
k = 479.
Hence, the required value of k is 479.
