There are no s's in a standard deck of cards, so the probability of selecting any s's, in any sequence of draws, in any strategy of replacement is exactly zero.
If only two cards are drawn from a standard deck of cards, with the first card replaced before drawing the second, the answer is 0.005917 (approx). If the first card is not replaced, the probability increases to 0.006033.
The probability is (4/52) * (4/52) = 1/169 = 0.0059, approx.
The probability of drawing a spade in a standard 52 card deck is 13 in 52, or 1 in 4. The probability of drawing a second spade, assuming the first spade was not replaced back into the deck, is 12 in 51. The probability, then, of drawing two spades is the product of those two probabilities, or 12 in 204, or 1 in 17.
When an event is repeated, the probability of it occurring is squared. For instance, if an outcome had the probability of 1/4, then the outcome happening twice would have a probability of 1/16. Note, however, that this does not mean that the second event has different probabilities. That particular outcome will always be 1/4, regardless of anything that happened before it.
Oh, dude, the probability of drawing 2 hearts from a deck of cards is like 1 out of 13 for the first card, and then 12 out of 51 for the second card. So, if you multiply those together, you get about a 4.5% chance of pulling off that heartwarming feat. But hey, who's counting, right?
The answer depends on whether or not the first card is replaced before the second is drawn.
The answer depends on whether the first number is replaced before picking the second. If not, the probability is 0.029
The answer depends on whether or not the first card is replaced before drawing the second.
If only two cards are drawn from a standard deck of cards, with the first card replaced before drawing the second, the answer is 0.005917 (approx). If the first card is not replaced, the probability increases to 0.006033.
2 in 52, or 1 in 26, or about 0.03846.
The answer depends on whether or not the first coin is replaced before choosing the second. Unfortunately, that critical information is not provided.
The probability is (4/52) * (4/52) = 1/169 = 0.0059, approx.
The answer depends onwhether the first letter is replaced before the second is selected, andwhether the order matters: are pa and ap equivalent.Order matters, no replacement: 1/30Order matters, first letter replaced: 1/36If the order does not matter, the probabilities are twice as large.
If you are drawing two cards from a full deck of cards (without jokers) then the probability will depend upon whether the the first card is replaced before the second is drawn, but the probability will also be different to being dealt a hand whilst playing Bridge (or Whist), which will again be different to being dealt a hand at Canasta. Without the SPECIFIC context of the two cards being got, I cannot give you a more specific answer.
The probability of rolling a 2 on the first roll is 1 in 6. The probability of rolling a 3 on the second roll is 1 in 6. However, the probability of rolling a 2 on the first roll and 3 on the second roll before you roll at all is 1/6 x 1/6 = 1 in 36.
Clearly, it is necessary to draw at least two cards. How many are drawn? Are the cards drawn at random? Is the first replaced before drawing the second? Please edit the question to include more context or relevant information.
The probability of drawing a spade in a standard 52 card deck is 13 in 52, or 1 in 4. The probability of drawing a second spade, assuming the first spade was not replaced back into the deck, is 12 in 51. The probability, then, of drawing two spades is the product of those two probabilities, or 12 in 204, or 1 in 17.