Two cards selected from standard deck of 52 cards. The first is replaced before second card selected. Find the probability of selecting a diamond and then a king
19/52
0.019
P(diamond)=13/52
P(king)=4/52
To select a diamond and THEN a king (with replacement) is then the product of the two above probabilities.
rounded to the nearest thousandth .193
Two cards are selected from a standard deck 52 playing cards. The first card it’s not replaced before the second card is selected. Find the probability of selecting a four and then selecting a two.
To find the probability of selecting a diamond and then a king from a standard deck of 52 cards, we need to consider the number of favorable outcomes (selecting a diamond and then a king) and divide it by the total number of possible outcomes.
First, let's determine the number of favorable outcomes:
1. Probability of selecting a diamond: There are 13 diamonds in a deck of 52 cards (one for each rank), so the probability of selecting a diamond is 13/52.
2. Since the first card is replaced before the second card is selected, the deck is effectively reset to its original composition. Therefore, the probability of selecting a king (there are 4 kings) as the second card is also 4/52.
Next, let's calculate the total number of possible outcomes:
Since we are selecting two cards with replacement, each card selection is independent. Therefore, the total number of possible outcomes is the product of the total number of outcomes for each card. For the first card, there are 52 possibilities, and for the second card, also 52 possibilities. So the total number of possible outcomes is 52 * 52 = 2704.
Finally, let's calculate the probability:
The probability of selecting a diamond and then a king is the product of the probabilities of each event occurring. Therefore, the probability is (13/52) * (4/52) = 1/52.
Hence, the probability of selecting a diamond and then a king from a standard deck of 52 cards, with replacement, is 1/52.