JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let h be the number of heads and t be the number of tails obtained. The total points obtained is 10h + 5t. For the man to get exactly 30 points, we must have: 10h + 5t = 30 2h + t = 6 Since h and t are non-negative integers, the possible pairs of (h, t) are (3, 0), (2, 2), (1, 4), and (0, 6). For each pair, the number of possible sequences of heads and tails is given by ^ h+t C_ h , and the probability of any specific sequence is ( 1 2 )^ h+t . The total probability is the sum of the probabilities of all these mutually exclusive cases: For h = 3, t = 0, the probability is ^ 3 C_ 3 ( 1 2 )^ 3 = 1 1 8 = 8 64 . For h = 2, t = 2, the probability is ^ 4 C_ 2 ( 1 2 )^ 4 = 6 1 16 = 24 64 . For h = 1, t = 4, the probability is ^ 5 C_ 1 ( 1 2 )^ 5 = 5 1 32 = 10 64 . For h = 0, t = 6, the probability is ^ 6 C_ 0 ( 1 2 )^ 6 = 1 1 64 = 1 64 . Total probability = 8 64 + 24 64 + 10 64 + 1 64 = 43 64 . Thus, m = 43 and n = 64. Since (43, 64) = 1, the fraction is in its simplest form. m + n = 43 + 64 = 107. Answer: 107