A money box contains four \( \$ 2 \) coins, seven \( \$ 1 \) coins, twelve 50 cent coins and twenty two 20 cent coins. Christie chooses a coin at random. Find the probability she chooses a coin that is either a \( \$ 1 \) or a \( \$ 2 \) coin.

To find the probability that Christie chooses either a \( \$ 1 \) coin or a \( \$ 2 \) coin, we first need to determine the total number of coins in the money box. The total number of coins is: - \( 4 \) (for the \( \$ 2 \) coins) - \( 7 \) (for the \( \$ 1 \) coins) - \( 12 \) (for the 50 cent coins) - \( 22 \) (for the 20 cent coins) Calculating the total: \[ 4 + 7 + 12 + 22 = 45 \] Now, we find the number of favorable outcomes, which is the total number of \( \$ 1 \) and \( \$ 2 \) coins: \[ 4 + 7 = 11 \] Now, the probability \( P \) of selecting either a \( \$ 1 \) coin or a \( \$ 2 \) coin is given by the formula: \[ P(\text{selecting a } \$ 1 \text{ or } \$ 2 \text{ coin}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of coins}} = \frac{11}{45} \] Therefore, the probability that Christie chooses a coin that is either a \( \$ 1 \) or a \( \$ 2 \) coin is \( \frac{11}{45} \).