 A balanced dice is tossed twice. Find the probability that it will show an even number in both the tosses.

A balanced die has six faces numbered 1 through 6, and among these, the even numbers are 2, 4, and 6. The probability of getting an even number in one toss of the die is:

\[
P(\text{even number}) = \frac{3}{6} = \frac{1}{2}
\]

Since the tosses are independent events, the probability of getting an even number on both tosses is the product of the individual probabilities:

\[
P(\text{even on both tosses}) = P(\text{even on first toss}) \times P(\text{even on second toss}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
\]

Thus, the probability that the dice will show an even number in both tosses is:

\[
\boxed{\frac{1}{4}}
\]

Would you like further details or have any questions? Here are 5 related questions you might want to explore:

1. What is the probability of getting odd numbers on both tosses?
2. How does the probability change if the dice are tossed three times?
3. What is the probability of getting at least one even number in two tosses?
4. Can we extend this to finding the probability of two specific numbers (like 2 and 4) appearing in two tosses?
5. What is the probability of getting the same even number on both tosses?

**Tip:** When calculating probabilities for independent events, always multiply the individual probabilities to find the combined outcome.

A balanced dice is tossed twice. Find the probability that it will show an even number in both the tosses.

Find the probability of getting an even number on both tosses of a balanced dice. This problem involves calculating probabilities for independent events using the multiplication rule, and is suitable for students in Grades 6-8.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation