To solve this probability question, we need to find the likelihood of obtaining an odd number at least once when a standard die is tossed three times. We can approach this by considering the complement event, which is often simpler for "at least once" scenarios.

Die Toss Outcomes Explained

A standard six-sided die has the following possible outcomes for a single toss:

• 1 (which is an odd number)
• 2 (which is an even number)
• 3 (which is an odd number)
• 4 (which is an even number)
• 5 (which is an odd number)
• 6 (which is an even number)

From these outcomes, we can determine the individual probabilities:

• Number of odd numbers = 3 (1, 3, 5)
• Number of even numbers = 3 (2, 4, 6)
• Total possible outcomes = 6

Probability for a Single Toss

For one toss of the die:

• The probability of getting an odd number, denoted as $P(\text{Odd})$: $$ P(\text{Odd}) = \frac{\text{Number of odd outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} $$
• The probability of getting an even number, denoted as $P(\text{Even})$: $$ P(\text{Even}) = \frac{\text{Number of even outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} $$

Calculating Complementary Probability for Three Tosses

The event "getting an odd number at least once" is the opposite (complement) of the event "getting no odd numbers at all". If there are no odd numbers in three tosses, it means all three tosses resulted in even numbers.

Since each die toss is an independent event (the result of one toss doesn't affect the others), we can multiply their individual probabilities to find the combined probability:

• Probability of getting an even number on the first toss: $P(\text{Even}_1) = \frac{1}{2}$
• Probability of getting an even number on the second toss: $P(\text{Even}_2) = \frac{1}{2}$
• Probability of getting an even number on the third toss: $P(\text{Even}_3) = \frac{1}{2}$

The probability of getting all three tosses as even numbers ($P(\text{All Even})$) is:

$$ P(\text{All Even}) = P(\text{Even}_1) \times P(\text{Even}_2) \times P(\text{Even}_3) $$ $$ P(\text{All Even}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} $$

Final Probability Result

Now, to find the probability of getting an odd number at least once, we use the complementary probability formula:

$$ P(\text{Event}) = 1 - P(\text{Complement of Event}) $$

Applying this to our problem:

$$ P(\text{at least one odd number}) = 1 - P(\text{all even numbers}) $$ $$ P(\text{at least one odd number}) = 1 - \frac{1}{8} $$

To perform the subtraction, we convert 1 to a fraction with a denominator of 8:

$$ P(\text{at least one odd number}) = \frac{8}{8} - \frac{1}{8} = \frac{7}{8} $$

Thus, the probability of getting an odd number at least once when a die is tossed three times is $\frac{7}{8}$.

Summary of Probabilities for Die Tosses

Event	Probability
Getting an odd number in one toss	$1/2$
Getting an even number in one toss	$1/2$
Getting all three tosses as even numbers	$1/8$
Getting at least one odd number in three tosses	$7/8$