1 / n!
The probability of the coin coming up heads each time is 1/8; likewise for 3 tails. The probability of getting 2 heads and 1 tail (in any order) or 2 tails and 1 head, is 3/8. There are lots of other events whose probability can be calculated when a coin is tossed 3 times, but the question doesn't specify what event is to have its probability calculated.
First, you find the probability of picking 1 in 5 disregarding order (combination). 5 C 1 = 5/1 = 5 Then you divide that by the possible number of outcomes which is 25. 5/ 2^5= .0195
If the order of the numbers matters, then:The probability of a 6 on the first roll is 1/6.The probability of a 1 on the second roll is 1/6.The probability of a 3 on the third roll is 1/6.The probability of all three is (1/6 x 1/6 x 1/6) = 1/216 = 0.00463 = 0.463 % (rounded)If you're not concerned about the order of the rolls, only the probability of getting all three numbers, then:The probability of getting any one of them on the first roll is 3/6, or 1/2.The probability of getting one of the other two on the second roll is 2/6, or 1/3.The probability of getting the final one on the third roll is 1/6.The chance of all three without order then is (1/2 x 1/3 x 1/6) = 1/36 or 2.778% (rounded)
Yes, you divide the number of expected outcomes by the number of possible outcomes in order to determine probability.
There are no numbers in alphabetical order! In alphanumeric order, numbers come first.
The probability of the coin coming up heads each time is 1/8; likewise for 3 tails. The probability of getting 2 heads and 1 tail (in any order) or 2 tails and 1 head, is 3/8. There are lots of other events whose probability can be calculated when a coin is tossed 3 times, but the question doesn't specify what event is to have its probability calculated.
First, you find the probability of picking 1 in 5 disregarding order (combination). 5 C 1 = 5/1 = 5 Then you divide that by the possible number of outcomes which is 25. 5/ 2^5= .0195
Assume the die is fair. What are the numbers on the four sides? How many times are you allowed to roll it? If the four numbers are 1,2,3 and 4 and you want the first three rolls to be 1,3,2 in that order, the probability is 1/4*1/4*1/4 = 1/64.
If the order of the numbers matters, then:The probability of a 6 on the first roll is 1/6.The probability of a 1 on the second roll is 1/6.The probability of a 3 on the third roll is 1/6.The probability of all three is (1/6 x 1/6 x 1/6) = 1/216 = 0.00463 = 0.463 % (rounded)If you're not concerned about the order of the rolls, only the probability of getting all three numbers, then:The probability of getting any one of them on the first roll is 3/6, or 1/2.The probability of getting one of the other two on the second roll is 2/6, or 1/3.The probability of getting the final one on the third roll is 1/6.The chance of all three without order then is (1/2 x 1/3 x 1/6) = 1/36 or 2.778% (rounded)
1/120there are three spots and six numbers on a diceso 6 numbers on the first spot and 5 numbers on the second spot and 4 in the third spot just multiply them
In order to calculate that, it seems to me that you'd have to know how many numbers exist all together, and how many pairs of numbers there are that sum to 10. Since both of these are infinite, I'll say that the calculation is not possible.
Assuming you want the probability FOR A SINGLE TRY, and you want the numbers in that exact order, the probability for each part (for instance, first = red; or second = green) is 1/4; therefore, the probability for the combination is (1/4) to the power 4.
If the sequence matters (you want H-H-T in that order), then . . .Probability of the first head = 0.5Probability of the second head = 0.5Probability of the tail on the 3rd toss = 0.5Probability of the correct 3-toss sequence = (0.5 x 0.5 x 0.5) = 1/8 = 0.125 = 12.5%=====================================================If the sequence doesn't matter, then the probability is higher.All the possible results of 3 tosses are:TTTTTHTHTTHHHTTHTHHHTHHHIf the sequence doesn't matter, there are 3 different ways to get 2 heads and 1 tail.The probability is3/8 = 0.375 = 37.5%
Yes, you divide the number of expected outcomes by the number of possible outcomes in order to determine probability.
There are no numbers in alphabetical order! In alphanumeric order, numbers come first.
it is the numbers 1-9 in alphabetical order!
The are in numerical order, not the order which they came out.