Problem 37 - Calculus for Biology and Medicine

Question

A loaded die is weighted so that rolling a 4 is three times as likely as rolling any of the other numbers. You roll the die twice and record the sum of the two numbers. What is the probability that the sum is equal to 7

Step-by-Step Solution

Answer

The probability that the sum is 7 is \( \frac{5}{32} \).

Step 1: Understand the die's probabilities

The die is weighted such that the probability of rolling a 4 is three times the probability of rolling any other number. Let the probability of rolling a number other than 4 be denoted by \( p \). The probability of rolling a 4 is then \( 3p \). The die has six faces, so the total probability must be 1: \( 5p + 3p = 1 \). Solve this equation to find \( p \): \( 8p = 1 \), hence \( p = \frac{1}{8} \). The probability for a 4 is thus \( \frac{3}{8} \). Now we have: \( P(1) = \frac{1}{8} \), \( P(2) = \frac{1}{8} \), \( P(3) = \frac{1}{8} \), \( P(4) = \frac{3}{8} \), \( P(5) = \frac{1}{8} \), \( P(6) = \frac{1}{8} \).

Step 2: Determine the combinations that sum to 7

To find the probability that the sum of two rolls is 7, we identify the combinations that result in 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Step 3: Calculate the probability for each combination

Calculate the probability for each pair: - \( P(1,6) = P(1) \cdot P(6) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \)- \( P(2,5) = P(2) \cdot P(5) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \)- \( P(3,4) = P(3) \cdot P(4) = \frac{1}{8} \cdot \frac{3}{8} = \frac{3}{64} \)- \( P(4,3) = P(4) \cdot P(3) = \frac{3}{8} \cdot \frac{1}{8} = \frac{3}{64} \)- \( P(5,2) = P(5) \cdot P(2) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \)- \( P(6,1) = P(6) \cdot P(1) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64} \)

Step 4: Sum the probabilities

Add up all the probabilities of the combinations that result in a sum of 7:\( \frac{1}{64} + \frac{1}{64} + \frac{3}{64} + \frac{3}{64} + \frac{1}{64} + \frac{1}{64} = \frac{10}{64} = \frac{5}{32} \).

Step 5: Conclusion

The probability that the sum of the two rolls is 7 is \( \frac{5}{32} \).

Key Concepts

Conditional Probability

Conditional probability helps us understand the likelihood of an event happening, given that another event has already occurred. It is an essential concept in probability theory that allows us to make more informed predictions. When dealing with a loaded or weighted die, such as in this exercise, conditional probability can play a role when trying to understand outcomes depending on specific conditions.

• Consider rolling a die twice. Rolling a sum of 7 involves specific outcomes based on each roll.
• If a roll results in a 4, it affects the overall probability of rolling a 7 on the second roll, since the probability of rolling a 4 is higher compared to other numbers.

This exercise doesn't involve explicit conditional probability calculations, but understanding conditional probabilities would enhance analyzing different types of probability problems.

Weighted Dice

A weighted or loaded die does not operate like a fair six-sided die. In our exercise, a certain number on this die is more likely to occur. Specifically, rolling a 4 is three times more likely than rolling any other number.

• With a weighted die, the probabilities assigned to each side aren't equal. For our die, the probability of rolling a 4 is higher due to its weighting.
• If we denote the probability of rolling any number other than 4 by \( p \), then the probability of 4 is \( 3p \).

To determine these probabilities, we know that the sum of probabilities of all possible outcomes in a dice roll must be 1. Therefore: 5 (other numbers) \( \times p \) + 1 (number 4) \( \times 3p = 1 \). Solving this gives us \( p = \frac{1}{8} \) and indicates that the probability of rolling a 4 is \( \frac{3}{8} \). Understanding weighted dice is crucial in problems where some outcomes are favored over others.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations of objects under specific conditions. In this exercise, combinatorics helps identify all possible outcomes when rolling two dice that sum up to 7.

• To find this probability, we first list all possible pairs of dice rolls: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).
• Each combination sums to 7 and needs to be considered for probability calculations.

Using combinatorics, we calculate the probability for each outcome, using the assigned probability values from our weighted die. Finally, we sum the probabilities for all valid combinations to determine the total probability of rolling a sum of 7.
This approach effectively organizes our problem-solving process, ensuring all possibilities are accounted for in determining the desired result.