Question

A die is thrown 10 times, the probability that an odd number will come up atleast one time is

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

Answer 1

The Correct Option is B

Let \(O\) be the event of getting an odd number when a die is thrown.

Let \(E\) be the event of getting an even number when a die is thrown.

The possible outcomes when a die is thrown are {1, 2, 3, 4, 5, 6}.

The odd numbers are {1, 3, 5}.

The even numbers are {2, 4, 6}.

The probability of getting an odd number is \(P(O) = \frac{3}{6} = \frac{1}{2}\).

The probability of getting an even number is \(P(E) = \frac{3}{6} = \frac{1}{2}\).

The die is thrown 10 times. We are looking for the probability that an odd number comes up at least once. It's easier to calculate the probability of the complement event (that no odd numbers come up, meaning only even numbers appear) and subtract it from 1.

The probability of getting an even number in one throw is \(P(E) = \frac{1}{2}\).

The probability of getting even numbers in all 10 throws is \(P(E)^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\).

The probability of getting at least one odd number is \(1 - P(\text{no odd numbers}) = 1 - \frac{1}{1024} = \frac{1024 - 1}{1024} = \frac{1023}{1024}\).

Thus, the probability that an odd number will come up at least one time is \(\frac{1023}{1024}\).