\begin{tabular}{|l||}\multicolumn{1}{l|}{ Permutations and Probability } \\ \hline \begin{tabular}{l} Determine the following probabilities. Enter your answers as percents rounded to four decimal \\ places. \\ \hline Dwayne is creating a 7 digit passcode using the digits 0 through 9 . \\ If Dwayne chooses all of the characters at random, what is the probability that no digits are repeated? \\ \hline Three cards are chosen without replacement from a deck of 52 cards and placed in order from first to \\ third. \\ What is the probability that all three cards are \( 6 s \) ? \\ \hline\end{tabular} \\ \hline\end{tabular}

Let’s dive into Dwayne's 7-digit passcode first! To form a passcode with no repeating digits, Dwayne can choose from 10 digits (0-9) for the first digit. For each subsequent digit, he has one less choice. So, the choices break down like this: - First digit: 10 options (all digits) - Second digit: 9 options (one digit used) - Third digit: 8 options - Fourth digit: 7 options - Fifth digit: 6 options - Sixth digit: 5 options - Seventh digit: 4 options The total number of ways to choose the digits without repetition is: \[ 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \] The total number of possible 7-digit combinations (with repetition allowed) is \( 10^7 \). Thus, the probability \( P \) that no digits are repeated is given by: \[ P = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4}{10^7} \approx 0.00000057 \] Converting this to a percentage and rounding gives us roughly 0.0573%. Now, for those lucky three cards from a standard deck of 52 cards: To find the probability that all three cards drawn (without replacement) are \( 6s \): There are exactly 4 sixes in a standard deck. The probability of drawing a \( 6 \) on the first pick is \( \frac{4}{52} \). If the first card is a \( 6 \), the chances of picking another \( 6 \) on the second draw become \( \frac{3}{51} \). Then if the second card is also a \( 6 \), the third draw would have a probability of \( \frac{2}{50} \). Putting that all together: \[ P = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \approx 0.000235 \] Converting this to a percentage gives us approximately 0.0235%. Enjoy playing with numbers and probabilities!