Chapter 11: Problem 13

A large number of regular six-sided dice are shaken together in a box, then dumped onto a table. Those showing 1 or 2 are removed, and the process is repeated with the remaining ones. (See Explore It Yourself \(11.1 .)\) Is the half-life of the dice greater than, equal to, or less than one throw?

Short Answer

Expert verified

Based on the step by step solution provided, determine if the half-life of the dice removal process is greater than, equal to, or less than one throw. Answer: The half-life of the dice removal process is greater than one throw. This is because it takes approximately 1.7095 rounds (or throws) for half the dice to remain in the process (not removed).

Step by step solution

Find the probability of a die being removed

In each round, a 6-sided die has a probability of 1/6 of showing each number, from 1 to 6. We are interested in the probability of showing either 1 or 2. The probability of a die being removed (showing 1 or 2) is the sum of the probabilities of showing a 1 or a 2: \(\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\).

Calculate the half-life

Half-life refers to the period in which the remaining quantity reduces by half. We want to find the number of rounds when the remaining dice are half of the initial amount. Let 'k' be the number of rounds needed for half the dice to be removed. Original number of dice = N₀ After k rounds, remaining dice = N₀ / 2 The probability of a die staying in the process (not being removed) after each round is (1 - probability of being removed): \( 1 - \frac{1}{3} = \frac{2}{3} \) After k rounds, the probability of a die still remaining in the process is \((\frac{2}{3})^{k}\). Considering the remaining dice after k rounds: N₀/2 = N₀ * \((\frac{2}{3})^{k}\) Let's solve this equation for k.

Solve the equation for k

We have the equation N₀/2 = N₀ * \((\frac{2}{3})^{k}\). Dividing both sides by N₀, we get: \(\frac{1}{2} = (\frac{2}{3})^{k}\) To solve for k, we will take the natural logarithm of both sides: \(ln(\frac{1}{2}) = ln((\frac{2}{3})^{k})\) Using the properties of logarithms, we get: \(k*ln(\frac{2}{3}) = ln(\frac{1}{2})\) Now, to find k, divide both sides by \(ln(\frac{2}{3})\): \(k = \frac{ln(\frac{1}{2})}{ln(\frac{2}{3})}= 1.7095\)

Analyze the value of k

Since k = 1.7095, it means that it takes approximately 1.7095 rounds (or throws) for half the dice to remain in the process (not removed). As a conclusion, the half-life of the dice is greater than one throw.

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A large number of regular six-sided dice are shaken together in a box, then dumped onto a table. Those showing 1 or 2 are removed, and the process is repeated with the remaining ones. (See Explore It Yourself \(11.1 .)\) Is the half-life of the dice greater than, equal to, or less than one throw?

Short Answer

Step by step solution

Find the probability of a die being removed

Calculate the half-life

Solve the equation for k

Analyze the value of k

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