The probability of rolling a Prime number on a standard 6-sided die is 3 in 6, or 0.5.
The sample space is [1 2 3 4 5 6] and the result space is [2 3 5]. 3 divided by 6 is 0.5.
2, 3 and 5 are prime so P = 0.5
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.
"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%.
It is 0.375
2, 3 and 5 are prime so P = 0.5
Zero. Since coins land on Heads or Tails and not numbers.
The probability is 1/6.
The probability of tossing an odd number (assumed on a die) is 3 in 6 or 1 in 2. The probability of tossing a tail (assumed on a coin) is 1 in 2. Since these are unrelated events, and the question said "and", simply multiply the probabilities to get 1 in 4.
the probability is 4 out of 6
It is 0.0833... recurring.
1/2 * 1/2 = 1/4 1/2= probability of landing an even number 1/2 = probability of landing a heads
Each time you toss the die the probability of rolling an even number is 3 out of 6 or 1/2. So, the probability of tossing three consecutive even numbers is (1/2)3 = 1/8 = 0.125, which is one chance in eight.
The probability of tossing a 1 or 2 on a six sided die is 2 in 6, or 1 in 3.
There are 36 possibilities when rolling two six sided die. 15 of them result in a prime number (2,3,5,7,11) So the probability is 15 in 36 or a 5 in 12 chance.
It is 2/3.