A lottery game has players selecting 4 numbers at random from the numbers \( 1-45 \). The state picks 20 winning numbers. If the player's 4 numbers are selected by the state, the player wins. What is the probability of winning? How many ways can the 4 numbers be chosen from the original 45 numbers, regardless of whether they are in the winning group or the losing group? \( \square \) ways (Type a whole number.) The probability of winning is \( \square \). (Round to four decimal places as needed.)

To find the total number of ways to choose 4 numbers from 45, we use the combination formula \( C(n, k) \), which is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] Here, \( n = 45 \) and \( k = 4 \): \[ C(45, 4) = \frac{45!}{4!(45-4)!} = \frac{45 \times 44 \times 43 \times 42}{4 \times 3 \times 2 \times 1} = 123,410 \] So, there are \( 123,410 \) ways to choose 4 numbers from 45. Next, to find the probability of winning, we need to determine how many ways the player's 4 numbers can be selected from the winning group of 20 numbers. The number of ways to select 4 winning numbers from the 20 is: \[ C(20, 4) = \frac{20!}{4!(20-4)!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4,845 \] Thus, the probability of winning is the number of favorable outcomes (winning combinations) over the total outcomes (all combinations): \[ P(\text{winning}) = \frac{C(20, 4)}{C(45, 4)} = \frac{4,845}{123,410} \approx 0.0392 \] So the complete answers are: Ways to choose 4 numbers: **123410** Probability of winning: **0.0392** (rounded to four decimal places)