hypergeometric distribution: f(k;N,n,m) = f(5;52,13,5)
There are 13 diamonds. Three cards are dealt. The probability of all of them being diamond is (13/52)(12/51)(11/50) = 1716/132600 = 11/850
There are 13 clubs in a standard deck of 52 cards. The probability, then, of drawing club is 13 in 52, or 1 in 4, or 0.25.
If the pack is well shuffled, the probability is 1/52.
The probability, if the cards are dealt often enough, is 1.On a single deal, the prob is 3.69379*10^-6
From a standard deck of playing cards, there are 52 cards. There are 13 cards for each Spades, Hearts, Clubs and Diamonds. For the first card, the probability of getting a Spade is 13/52. If you're going to replace the first card into the deck, then the probability of second card being a Spade is also 13/52 Therefore, 13/52 x 13/52 = 1/16 (one in every sixteen tries) If you're not going to replace the first card into the deck, the remaining deck will be left with a total of 51 cards and 12 Spades. Therefore the probability of second card being a Spade is 12/51 Lastly, 13/52 x 12/51 = 1/17 (one in every seventeen tries)
There are 13 diamonds. Three cards are dealt. The probability of all of them being diamond is (13/52)(12/51)(11/50) = 1716/132600 = 11/850
5
If the card is drawn randomly, the probability is 1/4.
As there are no 12 cards in a standard pack the probability is zero.
There are 13 clubs in a standard deck of 52 cards. The probability, then, of drawing club is 13 in 52, or 1 in 4, or 0.25.
The probability of the card being BETWEEN 8 and K is 4/13.
Since there are only four aces in a standard 52 card deck, the probability of being dealt five aces is zero.
If the pack is well shuffled, the probability is 1/52.
The probability is 0. One card cannot be a club and a spade!
Probability is desired options over total options. There are 6 faces on a standard dice, so NOT rolling a 5 is 5/6.
The probability, if the cards are dealt often enough, is 1.On a single deal, the prob is 3.69379*10^-6
There is one chance in four of the first card being a diamond (since there are four suits). There is one chance in four of the second card being a diamond. The chances of both being a diamond is 1/4*1/4 = 1/16, or thus one chance in 16