It is 0.4
The prime numbers from one to nine are 2, 3, 5, and 7. There are nine numbers from one to nine. The probability is 4 (the number of prime numbers) over 9 (the total number of numbers). Therefore, the probability of choosing a prime number is 4/9 or about 44 percent.
2, 3 and 5 are prime so P = 0.5
"The probability of getting a prime number in a die is 4/6" Actually there are 3 prime numbers on a die. 2, 3, and 5 are all prime numbers. So this tells you that you have 3 chances it will be a prime number and 3 chances it will not be a prime number. So the probability of getting a prime number on a die would be 3/6 or 1/2.
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 three prime numbers on the cube are: 2 3 and 5 so the probability is 3/6 or 1/2 simplified
The probability of selecting a 17 (or any number for that matter) is 1/20 or .05 or 5%.
In the range of 1 to 10, there are five odd numbers: 1, 3, 5, 7, and 9. Since there are a total of 10 numbers, the probability of selecting an odd number is the number of odd outcomes divided by the total outcomes. This gives us a probability of 5/10, which simplifies to 1/2 or 50%.
There are eight prime numbers between 1 and 20.2, 3, 5, 7, 11, 13, 17, 19If you randomly choose in number then you have an 8 in 20 chance of selecting a prime.The probability is selecting a prime number is 8/20 or 0.4
There are infinitely many numbers and so the probability of the second event is 0. As a result the overall probability is 0.
This cannot be answered Until and Unless a certain set of numbers are given as Sample Space.
50%, or 1 out of 2. There are 12 numbers, 6 even and 6 odd, so they have equal chance of being odd or even if one is selected
To find the probability of selecting a number from 20 to 30 that is divisible by 3, we first identify the numbers in that range: 21, 24, 27, and 30. There are four suitable candidates, so the probability of selecting one of them is 4 out of 11 (the total numbers from 20 to 30, inclusive). After replacing the selected number, we check which of these are divisible by 12. Among the numbers listed, only 24 is divisible by 12. Therefore, the probability of selecting a number divisible by 3 and then finding it divisible by 12 is 1 out of 11, which simplifies to approximately 0.0909 or 9.09%.
Zero, since there are infinitely more irrational numbers than rational numbers. Note that "zero probability" is not the same as "impossible" in this case. For more details, see the Wikipedia article on "Almost surely".
There are 20 numbers in total from 1 to 20. The even numbers in this range are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20, totaling 10 even numbers. Therefore, the probability of picking an even number is the number of even numbers divided by the total numbers, which is ( \frac{10}{20} = \frac{1}{2} ). Thus, the probability of selecting an even number is 0.5 or 50%.
from 1-25 there are 25 numbers 1 3 5 7 9 11 13 15 17 19 21 23 25 = 13 numbers [2 4 6 8 10 12 14 16 18 20 22 24 = 12 numbers] so probability of selecting an odd number is 13 in 25 13:25
30% chance. First, find the factors of 20: {1,2,4,5,10,20} There are 6 factors, out of 20 possible numbers. 6/20 = .3 = 30%
16 in 52 chance.