Math Problem Statement

A pair of dice is rolled ten times. Find the probability of rolling an odd total exactly once. (Round to three decimal places as needed.)

Solution

To solve this problem, let’s break it down systematically:


Given:

  • A pair of dice is rolled 10 times.
  • We are asked to find the probability of rolling an odd total exactly once.
  • We round the answer to three decimal places.

Step 1: Probability of rolling an odd total

When rolling two dice:

  1. Each die has 6 faces, so there are 6×6=366 \times 6 = 36 total outcomes.
  2. An odd total occurs when one die rolls odd and the other die rolls even. For example:
    • Die 1 is odd (1,3,51, 3, 5), and Die 2 is even (2,4,62, 4, 6).
    • Die 1 is even (2,4,62, 4, 6), and Die 2 is odd (1,3,51, 3, 5).

Each combination contributes 3×3=93 \times 3 = 9 outcomes. Hence, the total outcomes resulting in an odd sum are 9+9=189 + 9 = 18.

Thus, the probability of rolling an odd total on a single roll is: P(odd total)=1836=0.5P(\text{odd total}) = \frac{18}{36} = 0.5 Similarly: P(not odd total)=10.5=0.5P(\text{not odd total}) = 1 - 0.5 = 0.5


Step 2: Using the Binomial Distribution

Let XX represent the number of times an odd total is rolled in 10 trials. Since the probability of an odd total remains constant for each roll, XX follows a binomial distribution: P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k} where:

  • n=10n = 10 (number of trials),
  • k=1k = 1 (we want exactly 1 odd total),
  • p=0.5p = 0.5 (probability of an odd total).

Step 3: Plugging in the Values

The binomial coefficient is: (101)=10!1!(101)!=10\binom{10}{1} = \frac{10!}{1!(10-1)!} = 10

The probability becomes: P(X=1)=(101)(0.5)1(0.5)101P(X = 1) = \binom{10}{1} (0.5)^1 (0.5)^{10-1} P(X=1)=10(0.5)1(0.5)9=10(0.5)10P(X = 1) = 10 (0.5)^1 (0.5)^9 = 10 (0.5)^{10} P(X=1)=10×0.0009765625=0.009765625P(X = 1) = 10 \times 0.0009765625 = 0.009765625


Step 4: Rounding the Result

Rounding to three decimal places: P(X=1)=0.010P(X = 1) = 0.010


Final Answer:

The probability that an odd total is rolled exactly once is 0.010.


Would you like a deeper explanation of any part of this solution? Let me know! 😊

Here are 5 related questions:

  1. What is the probability of rolling an odd total exactly twice in 10 rolls?
  2. How does the binomial distribution apply to this situation?
  3. What if the number of trials was increased to 20 instead of 10?
  4. How can this problem be solved using a cumulative distribution function (CDF)?
  5. What is the probability of rolling an even total exactly once in 10 rolls?

Tip:

When using binomial probabilities, make sure the events are independent, and the probability of success (or failure) remains constant throughout the trials!

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Combinatorics

Formulas

Binomial Probability Formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Combinatorial Formula: C(n, k) = n! / (k!(n-k)!)

Theorems

Binomial Theorem

Suitable Grade Level

Grades 10-12