Math Problem Statement

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

Solution

The question asks:

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

Solution:

A standard balanced die has six faces with the numbers: 1, 2, 3, 4, 5, and 6. Out of these, the even numbers are: 2, 4, and 6.

  • The probability of rolling an even number on a single toss is the ratio of favorable outcomes (2, 4, 6) to the total number of outcomes (1, 2, 3, 4, 5, 6):

    P(even number)=36=12P(\text{even number}) = \frac{3}{6} = \frac{1}{2}

Since the die is tossed twice, and the two events (the first toss and the second toss) are independent, the probability of getting an even number on both tosses is:

P(even on both tosses)=P(even on 1st toss)×P(even on 2nd toss)=12×12=14P(\text{even on both tosses}) = P(\text{even on 1st toss}) \times P(\text{even on 2nd toss}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Final Answer:

The probability of rolling an even number on both tosses is 14\frac{1}{4}.


Would you like more details on how probability works or have any questions?

Here are some related questions you might find interesting:

  1. What is the probability of getting an odd number on both tosses?
  2. How would the probability change if the die were biased?
  3. What is the probability of getting at least one even number in two tosses?
  4. How do you calculate the probability of rolling a specific number twice?
  5. Can you calculate the probability of getting different combinations of even and odd numbers?

Tip: In independent events, the total probability is the product of individual event probabilities.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events
Basic Arithmetic

Formulas

P(even number) = 3/6 = 1/2
P(even on both tosses) = P(even on 1st toss) × P(even on 2nd toss)

Theorems

Multiplication Rule for Independent Events

Suitable Grade Level

Grades 7-9