Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following: \[ 10 \text { Pennies } 7 \text { Dimes } \] 18 Nickels 27 Quarters What is the probability that you reach into the jar and randomly grab a penny and then, without replacement, a dime? Express your answer as a fraction or a decimal number rounded to four decimal places.

To find the probability of grabbing a penny followed by a dime without replacement, we first need to determine the total number of coins in the jar. The jar contains: - 10 Pennies - 7 Dimes - 18 Nickels - 27 Quarters So, the total number of coins is: \[ 10 + 7 + 18 + 27 = 62 \text{ coins} \] The probability of grabbing a penny first is: \[ P(\text{Penny}) = \frac{\text{Number of Pennies}}{\text{Total Coins}} = \frac{10}{62} = \frac{5}{31} \] Once a penny is removed, there are now 61 coins left in total (62 - 1). The probability of then grabbing a dime is: \[ P(\text{Dime | Penny}) = \frac{\text{Number of Dimes}}{\text{Remaining Coins}} = \frac{7}{61} \] Now, we multiply these probabilities to find the overall probability of both events occurring: \[ P(\text{Penny and then Dime}) = P(\text{Penny}) \times P(\text{Dime | Penny}) = \frac{5}{31} \times \frac{7}{61} \] Calculating this gives: \[ P(\text{Penny and then Dime}) = \frac{5 \times 7}{31 \times 61} = \frac{35}{1891} \] To find the decimal representation rounded to four decimal places: \[ \frac{35}{1891} \approx 0.0185 \] Thus, the answer is: \[ \frac{35}{1891} \text{ or } 0.0185 \]