There are (99 - 9) = 90 2-digit numbers.
13 of them are multiples of 7.
The probability is 13/90 = 14.4% (rounded)
Given any random digit, there is a 10% probability that the next digit is the same. That is, the probability of the previous number repeating is 10%. Whether this constitutes often (as suggested by the question) or not is a matter of opinion.
49/9000
according to me its 1/2 as the possibility of getting an even is 1/2
The probability is 1/2.
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
Given any random digit, there is a 10% probability that the next digit is the same. That is, the probability of the previous number repeating is 10%. Whether this constitutes often (as suggested by the question) or not is a matter of opinion.
49/9000
It is 1/10.
according to me its 1/2 as the possibility of getting an even is 1/2
40/50
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
The probability is 1/2.
The probability is 1/2.
Divide the two-digit number by the one-digit number. If the remainder is zero then the 2-digit number is a multiple and if not, it is not.
It is 0.45
It is 4/9
To determine if a two-digit number is a multiple of a one-digit number, you can divide the two-digit number by the one-digit number. If the result is an integer (with no remainder), then the two-digit number is a multiple of the one-digit number. Alternatively, you can check if the two-digit number ends in a digit that confirms divisibility by the one-digit number (for example, even numbers for 2, ending in 0 or 5 for 5, etc.).