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How many 2's are in a deck of cards?
There are four 2s in a deck of cards.
How many combinations are there for a 5 digit code using 3 2s and 2 4s?
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
What is the probability of rolling two 2s followed by 1 on three tosses of a fair die?
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.
If you roll two fair dice once what is the probability you will obtain at least one 2 or one 5?
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.
You roll a pair of dice until you get doubles What is the probability that you get doubles in three or fewer rolls?
! 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
