NDA Mathematics Question (2025) — Solution

Step-by-step solution

Total number of possible outcomes when a fair coin is tossed 4 times is 2^4 = 16. Let E be the event that two heads do not occur consecutively. The favorable outcomes are the sequences of length 4 that do not contain 'HH'. We can count these by the number of heads in the sequence: Case 1: 0 heads TTTT (1 outcome) Case 2: 1 head HTTT, THTT, TTHT, TTTH (4 outcomes) Case 3: 2 heads To avoid consecutive heads, the heads must be separated by tails. The possible arrangements are HTHT, HTTH, THTH (3 outcomes) Case 4: 3 or 4 heads It is impossible to place 3 or 4 heads in 4 tosses without at least two heads being consecutive. (0 outcomes) Total number of favorable outcomes = 1 + 4 + 3 = 8. Required probability = 8 16 = 1 2 . Answer: 1/2