Q.A die is tossed thrice. Find the probability of getting an odd number at least once.

The key idea is the Probability Complement Rule: $P(\text{at least one}) = 1 - P(\text{none})$.

Step 1: On a single toss of a fair die, the odd numbers are 1, 3, 5. So $P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$, and $P(\text{not odd}) = 1 - \frac{1}{2} = \frac{1}{2}$.

Step 2: The die is tossed thrice independently. The probability of getting no odd number (i.e., an even number every time) is: …

The key insight here is that "at least once" is often easier to handle by thinking about its opposite: "never." When you see "at least one" in a probability problem, your first instinct should be to check if the complement is simpler to calculate. In this case, the complement — getting an even number on every toss — is a straightforward multiplication of independent probabilities.

A die has six faces: 1, 2, 3, 4, 5, 6. Odd numbers are 1, 3, 5 (three outcomes), and even numbers are 2, 4, 6 (three outcomes). So on a single toss, the probability of an odd number is $\frac{3}{6} = \frac{1}{2}$, and the probability of an even number is also $\frac{1}{2}$.

Since each toss is independent, the probability of getting an even number on all three tosses is simply the product of the individual probabilities.

1. Define the event of interest.

   Let $A$ be the event "getting an odd number at least once in three tosses." We want $P(A)$.

2. Identify the complement.

   The complement $A'$ is "getting no odd number in three tosses" — that is, every toss shows an even number.

3. Calculate the probability of the complement.

   On one toss, $P(\text{even}) = \frac{1}{2}$. For three independent tosses:

4. Apply the complement rule. …

Method: The "at least once" complement trick

Whenever a problem asks for the probability of getting some outcome at least once over several independent repetitions, computing it directly means adding many cases; the complement "it never happens" is a single clean case.

Steps

Step 1: Identify the single-trial success probability $p$ and its complement.

Find $p = P(\text{outcome on one trial})$, then the failure probability $q = 1 - p$.

Step 2: Multiply the failures across all independent trials.

For $n$ independent trials, "the outcome never occurs" needs a failure every time: …

Common Mistakes

Mistake 1: Trying to enumerate all "at least one odd" cases directly.

Why it's wrong: there are many overlapping cases (odd on one, two, or all three tosses), which is long and error-prone. Correct approach: use the complement, $P(\text{at least one})=1-P(\text{no odd})$.

Mistake 2: Taking $P(\text{no odd})$ as $\tfrac12$ instead of $\left(\tfrac12\right)^3$. …