An interesting question. The answer is 80%. A number must end in 0 or 5 to be dividable by 5. From 100 to 109 (10 numbers in total), there will be 2 numbers that are dividable by 5. If I then consider the entire set from 100 to 999 is 900 numbers, you have 90 sets of 10 numbers, each set with 2 dividable number, or 180/900 = 2/10 =20%. If 20% are dividable, 80% aren't. Suppose I extended the question, and say my set is all whole positive numbers and zero with n digits or less, where I pick n randomly from 1 to a million. If from this set, a number is picked randomly, what is the chance that it is dividable by 5. Answer: 20%.
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
It is possible to create a 3-digit number, without repeated digits so the probability is 1.
The question is ambiguous. It is in complete.
9/10
A three digit number can be chosen in 9*10*10 = 900ways No. of ways to choose a three digit number without 7 is 8*9*9 = 648 ways probability of picking a three digit number that includes atleast one digit that is 7 is 1- (648/900) = 252/900 = 7/25
49/9000
There is 100% chance.
90%
10,000,000 to one.
1 in 4,782,969
None of the digits can be 10, so the probability is 0.
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 is possible to create a 3-digit number, without repeated digits so the probability is 1.
Yes, yes, and no. 3- sum of digits must be multiple of 3. 6- sum of digits must be multiple of 3 and number must be even (multiple of 2). 9- sum of digits must be multiple of 9. (The sum of the digits here is 21.)
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
22