There are four 2s in a deck of cards.
There are five digits all together. The number of ways of arranging them is5 x 4 x 3 x 2 x 1 = 120 .But the three 2s can be arranged in 3 x 2 = 6 different ways and they all look the same.And the two 4s can be arranged in 2 different ways and they look the same.So out of the 120 different ways of arranging all five, there's a group of sixfor every possible arrangement of the 2s that all look the same, and there'sa group of two for each possible arrangement of the 4s that both look thesame.The number of combinations that look different is 120/(6 x 2) = 10 .And here they are:2 2 2 4 42 2 4 2 42 2 4 4 22 4 2 2 42 4 2 4 22 4 4 2 24 2 2 2 44 2 2 4 24 2 4 2 24 4 2 2 2
If you toss the die often enough then the probability of getting the sequence 2-2-1 is 1: a certainty. The probability of getting the result in the first three tosses is 1/216.
There are 6X6 or 36 different possibilities. Of these, 12 involve the first die being either a 2 or a 5 (as well as four results with either two 2s, two 5s or a 5 and a 2 (two of these, 5 on first 2 on second and 2 on first 5 on second)). With the 24 remaining rolls (where the first die is either 1,3,4 or 6), 2/6 will involve a 2 or a 5 on the second die. this gives us 8 new rolls in which the first die did not have a 2 or a 5 but the second did. So 12 + 8 gives us 20/36 or .555 cases where at least one of the rolls was either a 2 or a 5.An alternate way to get here is to realize that we have a 2/6 chance for each die to get a 2 or a 5. If we add these probabilities together we would conclude that there is a 4/6 probability and that 24 of the rolls will be either a 2 or a 5 BUT we have actually double-counted the cases where they BOTH have a 2 or a 5 (remember those 4 cases) so if you subtract those 4, it will get you to 20 as well.
! So first of all we need to calculate the probability of the base event, that is to get a pair when rolling 2 dice. This is quite simple and we can see it in 2 ways. a) We throw the first dice, regardless of what it comes up as, we have 1 in 6 chanses that the second dice matches it i.e probability = 1/6 b) There are 6*6 possible out comes when rolling two dice. Out of those we can get pair of 1s, pair of 2s , pair of 3s, pair of 4s, pair of 5s or pair of 6s. That is 6 pairs out of 36 total = 1/6 probability. Now about making a sequence of throws. The probability of us making the pair in the first throw is as the basic event = 1/6. If we make it we dont continue and have reached our goal! If we dont get a pair (proability 1-1/6 = 5/6) we continue and make our second throw. Again we have a 1/6 chanse, so in total that we make it in exactly the 2nd throws is 5/6 (miss)*1/6 (hit) = 5/36. If we dont make it , we continue again. Now our chance in making it in exactly the 3rd time is 5/6 (miss)*5/6 (miss again) *1/6 (hit) = 25/216. So the probability of making a pair in three or fewer rolls is the sum of the above, so 1/6 + 5/36 + 25/216 = 91/216 = 0.42 (roughly) Another way when you have established the probability of a single action to find how many repeats you need is to use the Binomial or Poisson distribution (look it up). Best Regards
There are four 2s in a deck of cards.
There are four 2s in a standard deck of playing cards.
There are four 2s in a deck of cards, one of each suit.
There are four 2s in a standard deck of playing cards.
There are four 2s in a standard deck of playing cards.
36 playing cards in a Russian deck. Russian deck starts with 6s. There are same four suits as American deck but 2s, 3s, 4s and 5s are missing.
Probability is 1/2 or 50%. Explanation: A normal deck of cards has 52 cards, consisting of four different suits each having 13 cards. Two suits are red, and two are black. This means the deck has 26 red cards. The probability that a red card will be drawn out of the 52 cards is 26 / 52 = 1/2.
In a traditional deck of cards, there are four of everything. Aces are the ones, so four Aces (Ace of Spades, Ace of Diamonds, Ace of Clubs, Ace of Hearts), four of each number between 2-10 (same suits), and four each of Jacks, Queens, and Kings. There are also two Jokers in a deck, although they are only used for certain games, and removed for others. Of course, if you are making your own, you can change the suits and change the royalty to types of ... cookies, or whatever, but that is what is in a traditional deck, so if you have the same amount of cards, you'll be able to play the same games.
It can be argued that a 'kind' of card is merely a particular grouping of the cards. Using this definition, there could be (omitting Jokers)... Red/Black - 2 Face/Non-face - 2 Club/Diamond/Heart/Spade - 4 1s,2s,3s,...Ks,As - 13 There could be additional possible groupings...Male/Female/Neuter, odd/even, divisible by 3/4/etc., cards with a face value <6... Then, we could combine some of these to create new groupings... Red Face/Red Non-face/Black Face/Black Non-face - 4 Color+suit is redundant, since suit is a subset of color Color+value (e.g. Black 1s, Red 1s,...) yields 26 categories Suit+Face/Non-face (e.g. Heart Face cards) - 8 Face/Non-face + Value is redundant Suit+value - 52 However, based on the vague definition of 'kinds of cards', there are too many possible solutions to solve this one.
There is only one 2s orbital in an atom.
25 of them
310.