Two cards are selected from a standard deck of 52 playing cards. The first card is not replaced before the second card is selected. Find the probability of selecting an eight and then selecting a seven.
Solution
Step 1 ：Define the total number of cards in a deck, which is 52, and the number of eights and sevens in the deck, which are both 4.
Step 2 ：Calculate the probability of selecting an eight from the deck. This is done by dividing the number of eights by the total number of cards. The result is \(\frac{4}{52} = 0.07692307692307693\).
Step 3 ：After selecting an eight, the card is not replaced, so there are now 51 cards left in the deck. Calculate the probability of selecting a seven from the remaining cards. This is done by dividing the number of sevens by the total number of remaining cards. The result is \(\frac{4}{51} = 0.0784313725490196\).
Step 4 ：The probability of both events happening is the product of their individual probabilities. So, the probability of selecting an eight and then a seven is \(0.07692307692307693 \times 0.0784313725490196 = 0.006033182503770739\).
Step 5 ：Final Answer: The probability of selecting an eight and then a seven from a standard deck of 52 playing cards, without replacement, is \(\boxed{0.006}\).
