Two fair coins are tossed. The probability of getting AT LEAST ONE head is
A$1/4$
B$1/2$
C$3/4$
D$1$
Answer & Solution
Correct answer: C. $3/4$
1. Sample space: $\{HH, HT, TH, TT\}$ — $|S| = 4$ equally likely outcomes.
2. Event 'at least one head': occurs when at least one of the two coins shows H. Enumerate: $\{HH, HT, TH\}$ — three outcomes.
3. $P(\text{at least 1 H}) = 3/4$.
4. Alternative via COMPLEMENT: $P(\text{at least 1 H}) = 1 - P(\text{no H}) = 1 - P(\text{TT}) = 1 - 1/4 = 3/4$. (Often faster.)
5. Option A is $P(\text{exactly two heads}) = 1/4$. Option B is $P(\text{exactly one head}) = 1/2$. Option D would mean certainty.
_Source: NCERT Class 11 Mathematics, Ch 14, §14.2 (Events and probability calculation), p. 4–5._
Related questions
Two coins tossed. P(both heads) =Bag has 4 red, 6 blue. P(red) =Probability of rolling 6 on a die:Probability of heads on a fair coin is:If a fair six-sided die is rolled twice, the probability that the sum is divisible by 3 isA bag has 3 red and 2 black balls. Probability of red ball drawn at random:Probability of getting an even number on rolling a single die:Probability of drawing a king from a standard deck of 52 cards: