Afair coin is tossed 5 times. what is the probability of obtaining no more than two heads?

1/2 This problem can be easily solved 2 ways, you can enumerate all 32 possibilities and count the number that have 0, 1, or 2 heads. Or you can calculate result directly. For no more than 2 heads, we want the sum of 0 heads, 1 head, and 2 heads. So 0 heads = (1/2)^5 = 1/32 1 head = (1/2)*(1/2)^5 * 5!/(1!4!) = 1/32 * 5 = 5/32 2 heads = (1/2)^2*(1/2)^5 * 5!/(2!3!) = 1/32 * 10 = 10/32 And finally: 1/32 + 5/32 + 10/32 = 16/32 = 1/2