Assuming then that there are 100 numbers, 1-100, the probability of the number 23 being randomly picked out of 100 is: 1/100 or 0.01.
15/49ExplanationThere are 15 prime numbers between 1 and 49 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47). If you randomly choose one natural number from the 49 numbers between 1 and 49 inclusive, there is a 15/49 probability that it will be prime.
If the die is fair then for a single roll, the probability is 1/2.
2/6 or 1/3
There are infinitely many numbers and so the probability of the second event is 0. As a result the overall probability is 0.
50 50 odd or even same probability
It is 0.02
The probability that a randomly chosen student is a woman can be calculated by dividing the number of women by the total number of students in the class. In this case, there are 13 women and 31 total students, so the probability is 13/31, which simplifies to approximately 0.419 or 41.9%.
1/3
There are 12 composite (and 8 primes) in the first twenty whole numbers. So the probability of randomly choosing a non-prime is 12/20 or 60%.
the first 10 whole numbers are numbers 1 to 10 and in those numbers only 3 numbers are divisible by 3 in which 3, 6 and 9 therefore the probability of from those figures that the numbers won't be divisible by 3 is 7/10 or 70%.
Assuming then that there are 100 numbers, 1-100, the probability of the number 23 being randomly picked out of 100 is: 1/100 or 0.01.
50%
Assuming the uniform continuous distribution, the answer is 29/49. With the uniform discrete distribution, the answer is 29/50.
The probability is 67/200.
Total number of woman over total number of people So for this problem it's 23/41
The answer is 4/9 or 44.444... % Any permutation of the digits 1 to 9 will be divisible by 9. So divisibility by 18 depends only on whether or not the last digit is even. If the last digit is 2, 4, 6 or 8 the number is divisible by 18 and if it is 1, 3, 5, 7 or 9 it is not. So 4 favourable outcomes out of 9 ie probability = 4/9.