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 3/4 and the remaining 6 questions correctly with probability 1/4. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is 27k/4¹⁰, then k is equal to ___________.

Solution

Maths sir · 3 steps

Exactly 8 right means exactly 2 wrong, so share those 2 wrong answers between the two groups in every way.

1. Split the wrong answers. Call the 4 questions with p = 3/4 group A and the 6 with p = 1/4 group B. Exactly 8 correct means exactly 2 wrong, split as (0 in A, 2 in B), (1, 1) or (2, 0). Use the binomial formula in each group and multiply, because the groups are independent.

   Wrong in A, B	Group A (4 questions)	Group B (6 questions)	Product × 4¹⁰
   0, 2	(3/4)⁴ = 81/4⁴	C(6, 2)(1/4)⁴(3/4)² = 135/4⁶	81 × 135 = 10935
   1, 1	C(4, 1)(3/4)³1/4 = 108/4⁴	C(6, 1)(1/4)⁵3/4 = 18/4⁶	108 × 18 = 1944
   2, 0	C(4, 2)(3/4)²(1/4)² = 54/4⁴	(1/4)⁶ = 1/4⁶	54 × 1 = 54

2. Add the cases. The total probability is (10935 + 1944 + 54)/4¹⁰ = 12933/4¹⁰. Taking out 27 gives 12933 = 27 × 479.
3. So. 27k/4¹⁰ = (27 × 479)/4¹⁰, so k = 479.

The common slip: 405. Counting only the case where all 4 easy questions are right and 2 of the other 6 are wrong gives 10935/27 = 405. The (1, 1) and (2, 0) splits are missing.

The idea to keep. When questions have different success probabilities, fix how many failures fall in each group, use the binomial formula group by group, multiply, then add over all splits.

Answer from the NTA final answer key. Solution by PrepWiser.