AMU 2026 — Mathematics PYQ

When three coins are tossed simultaneously, the total number of possible outcomes is $2^3=8$. The sample space is:

$$S=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$$

To find the probability of getting "at least one head," it is much easier to use the Complement Rule. The complement of "at least one head" is "zero heads" (i.e., getting all tails).

1. Identify the number of unfavorable outcomes:

   There is only one outcome where no heads appear: $\{TTT\}$.

   So, $P(\text{no heads})=1/8$.

2. Apply the complement rule:

   The sum of the probability of an event and its complement is always $1$.

   $$P(\text{at least one head})=1-P(\text{no heads})$$

   $$P(\text{at least one head})=1-\frac{1}{8}$$

   $$P(\text{at least one head})=\frac{7}{8}$$

Would you like me to explain how the sample space would change if we were tossing four coins instead of three?