Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is:
A. 1/16
B. 1/4
C. 7/8
D. 15/16
GATE 2002 · Engineering Mathematics · Binomial Distribution · medium
Answer: P(at least one head and at least one tail) = 7/8.
Count total outcomes and the complement: Total outcomes = 2^4 = 16. The complement event (NOT having at least one H and one T) consists of: all heads (HHHH) = 1 outcome, all tails (TTTT) = 1 outcome. So complement has 2 outcomes.
Apply the complement rule: P(all H) = 1/16. P(all T) = 1/16. P(at least one H and one T) = 1 - 1/16 - 1/16 = 1 - 2/16 = 14/16 = 7/8.