A radio station runs a promotion at an auto show with a money box with \( 13 \$ 50 \) tickets, \( 13 \$ 25 \) tickets, and \( 15 \$ 5 \) tickets. The box contains an additional 20 "dummy" tickets with no value. Three tickets are randomly drawn. Find the probability that exactly two \( \$ 50 \) prizes and no other money winners are choser The probability that exactly two \( \$ 50 \) prizes and no other money winners are chosen is \( \square \). (Round to four decimal places as needed.)

To find the probability that exactly two $50 prizes and no other money winners are chosen, we need to calculate the total number of ways to choose 3 tickets out of the total number of tickets and then find the number of ways to choose exactly two $50 prizes and no other money winners. Given: - Number of $50 tickets = 13 - Number of $25 tickets = 13 - Number of $5 tickets = 15 - Number of dummy tickets = 20 Total number of tickets = 13 + 13 + 15 + 20 = 61 We need to choose 3 tickets out of 61. This can be calculated using the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where: - \( n \) is the total number of items - \( k \) is the number of items to choose In this case, \( n = 61 \) and \( k = 3 \). \[ C(61, 3) = \frac{61!}{3!(61-3)!} \] Now, we need to find the number of ways to choose exactly two $50 prizes and no other money winners. This can be calculated by choosing 2 $50 tickets out of 13 and 1 dummy ticket out of 20. \[ C(13, 2) \times C(20, 1) \] Let's calculate these values and then find the probability. Simplify the expression by following steps: - step0: Solution: \(\frac{61!}{\left(3!\times \left(61-3\right)!\right)}\) - step1: Remove the parentheses: \(\frac{61!}{3!\times \left(61-3\right)!}\) - step2: Subtract the numbers: \(\frac{61!}{3!\times 58!}\) - step3: Expand the expression: \(\frac{61\times 60\times 59\times 58!}{3!\times 58!}\) - step4: Reduce the fraction: \(\frac{61\times 60\times 59}{3!}\) - step5: Multiply the terms: \(\frac{215940}{3!}\) - step6: Calculate the value: \(\frac{215940}{6}\) - step7: Reduce the fraction: \(35990\) Calculate or simplify the expression \( 13!/(2!*(13-2)!) \). Simplify the expression by following steps: - step0: Solution: \(\frac{13!}{\left(2!\times \left(13-2\right)!\right)}\) - step1: Remove the parentheses: \(\frac{13!}{2!\times \left(13-2\right)!}\) - step2: Subtract the numbers: \(\frac{13!}{2!\times 11!}\) - step3: Expand the expression: \(\frac{13\times 12\times 11!}{2!\times 11!}\) - step4: Reduce the fraction: \(\frac{13\times 12}{2!}\) - step5: Multiply the numbers: \(\frac{156}{2!}\) - step6: Calculate the value: \(\frac{156}{2}\) - step7: Reduce the fraction: \(78\) Calculate or simplify the expression \( 20!/(1!*(20-1)!) \). Simplify the expression by following steps: - step0: Solution: \(\frac{20!}{\left(1!\times \left(20-1\right)!\right)}\) - step1: Remove the parentheses: \(\frac{20!}{1!\times \left(20-1\right)!}\) - step2: Subtract the numbers: \(\frac{20!}{1!\times 19!}\) - step3: Reduce the fraction: \(\frac{20!}{19!}\) - step4: Expand the expression: \(\frac{20\times 19!}{19!}\) - step5: Reduce the fraction: \(20\) The total number of ways to choose 3 tickets out of 61 is \( C(61, 3) = 35990 \). The number of ways to choose exactly two $50 prizes and no other money winners is \( C(13, 2) \times C(20, 1) = 78 \times 20 = 1560 \). Now, we can find the probability that exactly two $50 prizes and no other money winners are chosen by dividing the number of favorable outcomes by the total number of outcomes. \[ P(\text{exactly two $50 prizes and no other money winners}) = \frac{1560}{35990} \] Let's calculate this probability. Calculate the value by following steps: - step0: Calculate: \(\frac{1560}{35990}\) - step1: Reduce the fraction: \(\frac{156}{3599}\) The probability that exactly two $50 prizes and no other money winners are chosen is approximately 0.0433 (rounded to four decimal places).