Well, honey, there are 90 two-digit numbers between 1 and 100, and a total of 100 numbers in that range. So, the probability of picking a two-digit number is 90/100, which simplifies to 9/10. So, the chances are pretty darn high, darling!
49/9000
1 in 4,782,969
The answer is 9*9!/9*109 = 0.0003629 approx.
It depends on how the game is structured, a piece of information that was not given in the answer... If the numbers are reused, then there are 1 x 107 (ten billion, in the short scale) permutations. Without combinations, then, the probability is 1 x 10-7. If the numbers are not reused, and we are playing combinatorial, in the style of most state sponsored lotteries, then the number of combinations is (10-3)! / 7!, which is 120, making the probability 1 in 120, or about 0.00833. There are other possibilities but, again, it depends on the rules, and you did not specify them completely.
The first (hundreds) digit can be chosen in any one of 9 ways from the digits {1,2,3,...,9}. With each one of these, the second (tens) digit can be chosen in any one of 10 ways from the digits {0,1,2,3,...,9}. With each of these combinationsm the third (units) digit can be chosen in one of 5 ways from the digits {1,3,5,7,9}. Altogether, then, there are 9*10*5 = 450 ways or 450 odd 3-digit numbers.
There is 100% chance.
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
49/9000
10,000,000 to one.
1 in 4,782,969
None of the digits can be 10, so the probability is 0.
you forgot the last 2 digits of your user ID for a games website. You know that both digits are odd. Find the probability that you type the correct last digits by randomly typing 2 odd numbers
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
The answer is 9*9!/9*109 = 0.0003629 approx.
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
It depends on how the game is structured, a piece of information that was not given in the answer... If the numbers are reused, then there are 1 x 107 (ten billion, in the short scale) permutations. Without combinations, then, the probability is 1 x 10-7. If the numbers are not reused, and we are playing combinatorial, in the style of most state sponsored lotteries, then the number of combinations is (10-3)! / 7!, which is 120, making the probability 1 in 120, or about 0.00833. There are other possibilities but, again, it depends on the rules, and you did not specify them completely.
The first (hundreds) digit can be chosen in any one of 9 ways from the digits {1,2,3,...,9}. With each one of these, the second (tens) digit can be chosen in any one of 10 ways from the digits {0,1,2,3,...,9}. With each of these combinationsm the third (units) digit can be chosen in one of 5 ways from the digits {1,3,5,7,9}. Altogether, then, there are 9*10*5 = 450 ways or 450 odd 3-digit numbers.