A standard 52 -card playing deck contains 4 of each type of card. Damon is dealt 7 cards from a fully shuffied deck. What is the probability his hand contains exactly 2 aces? \( \frac{594}{1,105} \)

To find the probability that Damon's hand contains exactly 2 aces when he is dealt 7 cards from a fully shuffled deck, we can use the concept of combinations. Given: - Total number of cards in the deck = 52 - Number of aces in the deck = 4 - Number of cards dealt to Damon = 7 We need to find the probability that Damon's hand contains exactly 2 aces. The total number of ways to choose 7 cards from a deck of 52 is given by the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where: - \( n \) is the total number of items (52 cards in this case) - \( k \) is the number of items to choose (7 cards in this case) The number of ways to choose 2 aces from the 4 aces in the deck is also given by the combination formula: \[ C(4, 2) = \frac{4!}{2!(4-2)!} \] The number of ways to choose the remaining 5 cards from the non-ace cards in the deck is given by the combination formula: \[ C(48, 5) = \frac{48!}{5!(48-5)!} \] The probability of Damon's hand containing exactly 2 aces is the ratio of the number of favorable outcomes (hands with exactly 2 aces) to the total number of possible outcomes (all possible hands of 7 cards). Therefore, the probability is: \[ P(\text{exactly 2 aces}) = \frac{C(4, 2) \times C(48, 5)}{C(52, 7)} \] Now, let's calculate the values of the combinations and the probability. Simplify the expression by following steps: - step0: Solution: \(\frac{\frac{\frac{4!}{\left(2!\times \left(4-2\right)!\right)}\times 48!}{\left(5!\times \left(48-5\right)!\right)}}{\left(\frac{52!}{\left(7!\times \left(52-7\right)!\right)}\right)}\) - step1: Remove the parentheses: \(\frac{\frac{\frac{4!}{2!\times \left(4-2\right)!}\times 48!}{5!\times \left(48-5\right)!}}{\frac{52!}{7!\times \left(52-7\right)!}}\) - step2: Subtract the numbers: \(\frac{\frac{\frac{4!}{2!\times 2!}\times 48!}{5!\times \left(48-5\right)!}}{\frac{52!}{7!\times \left(52-7\right)!}}\) - step3: Subtract the numbers: \(\frac{\frac{\frac{4!}{2!\times 2!}\times 48!}{5!\times 43!}}{\frac{52!}{7!\times \left(52-7\right)!}}\) - step4: Subtract the numbers: \(\frac{\frac{\frac{4!}{2!\times 2!}\times 48!}{5!\times 43!}}{\frac{52!}{7!\times 45!}}\) - step5: Reduce the fraction: \(\frac{\frac{\frac{4\times 3}{2!}\times 48!}{5!\times 43!}}{\frac{52!}{7!\times 45!}}\) - step6: Multiply the numbers: \(\frac{\frac{\frac{12}{2!}\times 48!}{5!\times 43!}}{\frac{52!}{7!\times 45!}}\) - step7: Divide the terms: \(\frac{\frac{6\times 48!}{5!\times 43!}}{\frac{52!}{7!\times 45!}}\) - step8: Reduce the fraction: \(\frac{\frac{6\times 48\times 47\times 46\times 45\times 44}{5!}}{\frac{52!}{7!\times 45!}}\) - step9: Multiply the terms: \(\frac{\frac{1232858880}{5!}}{\frac{52!}{7!\times 45!}}\) - step10: Calculate: \(\frac{\frac{1232858880}{5!}}{\frac{52!}{5040\times 45!}}\) - step11: Divide the terms: \(\frac{10273824}{\frac{52!}{5040\times 45!}}\) - step12: Divide the terms: \(\frac{10273824}{133784560}\) - step13: Reduce the fraction: \(\frac{594}{7735}\) The probability that Damon's hand contains exactly 2 aces when he is dealt 7 cards from a fully shuffled deck is \( \frac{594}{7735} \) or approximately 0.076794.