Question

A die is thrown 10 times. The probability that an odd number will come up at least once is:

• A. $\frac{11}{1024}$
• B. $\frac{1013}{1024}$
• C. $\frac{1023}{1024}$
• D. $\frac{1}{1024}$

Answer 1

The Correct Option is C

1. Understand the problem:

We need the probability that at least one odd number appears when a die is thrown 10 times.

2. Calculate the probability of the complement event:

The probability of no odd numbers (i.e., all even numbers) in 10 throws:

\[ P(\text{all even}) = \left(\frac{3}{6}\right)^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \]

3. Compute the desired probability:

\[ P(\text{at least one odd}) = 1 - P(\text{all even}) = 1 - \frac{1}{1024} = \frac{1023}{1024} \]

4. Match the result to the options:

The probability \( \frac{1023}{1024} \) corresponds to option (C).

Correct Answer: (C) \( \frac{1023}{1024} \)

Answer 2

A standard die has 6 faces ($ 1, 2, 3, 4, 5, 6 $). There are 3 odd numbers ($ 1, 3, 5 $) and 3 even numbers ($ 2, 4, 6 $).

The probability of getting an even number (not odd) on a single throw is:

$$ \frac{3}{6} = \frac{1}{2}. $$

Since each throw is independent, the probability of getting an even number on all 10 throws is $ \left(\frac{1}{2}\right) $ multiplied by itself 10 times:

$$ \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}. $$

This is the complement of getting all even numbers. Subtract the probability of getting all even numbers from 1:

$$ 1 - \frac{1}{1024} = \frac{1023}{1024}. $$

The probability of getting at least one odd number in 10 throws is $ \frac{1023}{1024} $. 

Therefore, the answer is $ \boxed{\text{(C)}} $.