JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let the outcomes of the 8 tosses be denoted by X_1, X_2, X_3, X_4, X_5, X_6, X_7, X_8. Let A be the number of heads in the first 3 tosses (X_1, X_2, X_3). Let B be the number of heads in the next 3 tosses (X_4, X_5, X_6). Let C be the number of heads in the last 2 tosses (X_7, X_8). According to the given conditions: Number of heads in the first six tosses: A + B = 4 Number of heads in the last five tosses: B + C = 3 Since C is the number of heads in 2 tosses, C 2. Thus, B = 3 - C 1. Also, B 3. The possible values for B are 1, 2, 3. We analyze each case: Case 1: B = 1 Then C = 2 and A = 3. Number of ways = ^ 3 C_ 3 ^ 3 C_ 1 ^ 2 C_ 2 = 1 3 1 = 3 Case 2: B = 2 Then C = 1 and A = 2. Number of ways = ^ 3 C_ 2 ^ 3 C_ 2 ^ 2 C_ 1 = 3 3 2 = 18 Case 3: B = 3 Then C = 0 and A = 1. Number of ways = ^ 3 C_ 1 ^ 3 C_ 3 ^ 2 C_ 0 = 3 1 1 = 3 Total number of favorable outcomes = 3 + 18 + 3 = 24 Total number of possible outcomes for 8 coin tosses = 2^8 = 256 The probability p is given by: p = 24 256 = 3 32 We need to find the value of 96p: 96p = 96 3 32 = 3 3 = 9 Answer: 9