Works for all numbers possible
The answer depends on what the twenty numbers are!
Although there are infinitely many primes, they become rarer and rarer so that as the number of numbers increases, the probability that picking one of them at random is a prime number tends to zero*. In the first 10 numbers there are 4 primes, so the probability of picking one is 4/10 = 2/5 = 0.4 In the first 100 numbers there are 26 primes, so the probability of picking one is 25/100 = 1/4 = 0.25 In the first 1,000 numbers there are 169 primes, so the probability of picking one is 168/1000 = 0.168 In the first 10,000 numbers there are 1,229 primes, so the probability of picking one is 0.1229 In the first 100,000 numbers there are 9592 primes, so the probability of picking one is 0.09592 In the first 1,000,000 numbers there are 78,498 primes, so the probability of picking one is 0.078498 In the first 10,000,000 numbers there are 664,579 primes, so the probability of picking one is 0.0664579 * Given any small value ε less than 1 and greater than 0, it is possible to find a number n such that the probability of picking a prime at random from the numbers 1-n is less than the given small value ε.
By picking numbers and graphing it loser! hehehehhe
The answer depends on whether the first number is replaced before picking the second. If not, the probability is 0.029
The probability is 1/b.
The probability is 11/21.
Depending on what numbers are you picking from: {Integers, Whole Numbers, Natural numbers, All real numbers} will affect the probability.
Just answer the problem by actually picking numbers from the solution and doing it.
That all depends on how many numbers you pick them out of.
The probability of picking a distinct set of 3 numbers from 20 is1/[20!/(3!)(17!)]= 1/1140The probability of only picking 3 from 20 is1/20
Here are some good ones: apple picking adding numbers acting
The probability is 8/20.