The table shows the number of coins in a change dish. If Malcolm selects two coins at random without replacement, what is the probability that he selects a nickel followed by a dime? Express your answer as a decimal rounded to the nearest thousandth if necessary.

\begin{tabular}{|c|c|}
\hline Coin & Number \\
\hline Penny & 8 \\
\hline Nickel & 6 \\
\hline Dime & 8 \\
\hline Quarter & 7 \\
\hline
\end{tabular}

- Calculate the total number of coins: \(8 + 6 + 8 + 7 = 29\). - Determine the probability of selecting a nickel first: \(\frac{6}{29}\). - Determine the probability of selecting a dime second, given a nickel was already removed: \(\frac{8}{28}\). - Multiply these probabilities and round to the nearest thousandth: \(\frac{6}{29} \times \frac{8}{28} \approx \boxed{0.059}\). ### Explanation 1. Understand the problem and provided data We are given a change dish containing different types of coins: \(8\) pennies, \(6\) nickels, \(8\) dimes, and \(7\) quarters. Malcolm selects two coins at random without replacement. Our goal is to find the probability that he selects a nickel first, followed by a dime, expressed as a decimal rounded to the nearest thousandth. 2. Calculate the total number of coins First, let's determine the total number of coins in the dish. We sum the number of each type of coin: \[\text{Total coins} = \text{Pennies} + \text{Nickels} + \text{Dimes} + \text{Quarters}\] \[\text{Total coins} = 8 + 6 + 8 + 7 = 29\] So, there are \(29\) coins in total in the dish. 3. Calculate the probability of selecting a nickel first The probability of selecting a nickel on the first draw is the number of nickels divided by the total number of coins. There are \(6\) nickels and \(29\) total coins. \[P(\text{first coin is a nickel}) = \frac{\text{Number of nickels}}{\text{Total number of coins}} = \frac{6}{29}\] 4. Calculate the probability of selecting a dime second After Malcolm selects a nickel, there are now \(28\) coins remaining in the dish (since the selection is without replacement). The number of dimes remains \(8\) because a nickel was removed, not a dime. So, the probability of selecting a dime on the second draw, given that a nickel was selected first, is: \[P(\text{second coin is a dime} \mid \text{first coin was a nickel}) = \frac{\text{Number of dimes}}{\text{Remaining total coins}} = \frac{8}{28}\] 5. Calculate the combined probability To find the probability of both events happening in sequence (selecting a nickel first AND then a dime), we multiply the probabilities of the individual events: \[P(\text{nickel then dime}) = P(\text{first coin is a nickel}) \times P(\text{second coin is a dime} \mid \text{first coin was a nickel})\] \[P(\text{nickel then dime}) = \frac{6}{29} \times \frac{8}{28}\] \[P(\text{nickel then dime}) = \frac{48}{812}\] Now, we convert this fraction to a decimal and round it to the nearest thousandth: \[\frac{48}{812} \approx 0.0591133...\] Rounding to the nearest thousandth, we get \(0.059\). Therefore, the probability that Malcolm selects a nickel followed by a dime is approximately \(0.059\). ### Examples Understanding sequential probabilities is crucial in many real-world scenarios, such as quality control in manufacturing. For instance, if you're inspecting a batch of \(29\) products with \(6\) known defects and \(8\) minor flaws, calculating the probability of picking a defective item first and then one with a minor flaw helps in assessing the overall quality and identifying potential issues in the production line. This ensures that the chances of encountering specific combinations of issues are well understood, allowing for better process adjustments.