+489 Two different integers from 1 through 20 inclusive are chosen at random. What is the probability that both numbers are prime? Express your answer as a common fraction.
The prime numbers between 1 - 20 are 2, 3, 5, 7, 11, 13, 17, and 19. There are 8 prime numbers.
On the first draw, the chances are 8/20 since there are 8 prime numbers and 20 total numbers. This can be simplified to 2/5.
On the second draw, the chances of a prime number are 7/19 if we got a prime number on the first draw. This is because one prime number has been used already and one number from the 20 total numbers has been used already (the problem says two *different* integers are chosen so we cannot pick the same number again). So there are 7 prime numbers and 19 total numbers.
The chances of getting a prime number both times is 7/19 of 2/5. To get the answer in one fraction, we multiply (7/19)*(2/5) = 14/95.
The answer is 14/95.
9/20 of those are prime, and you have double chances so your answer is 18/40 (Given these are normal people choosing random numbers)
There are 8 prime numbers between 1 - 20.
Therefore the odds are: 8/20 x 8/20 =64/400 =0.16 or 16%, assuming that 2nd number is drawn from 20 numbers.
The prime numbers between 1 - 20 are 2, 3, 5, 7, 11, 13, 17, and 19. There are 8 prime numbers.
On the first draw, the chances are 8/20 since there are 8 prime numbers and 20 total numbers. This can be simplified to 2/5.
On the second draw, the chances of a prime number are 7/19 if we got a prime number on the first draw. This is because one prime number has been used already and one number from the 20 total numbers has been used already (the problem says two *different* integers are chosen so we cannot pick the same number again). So there are 7 prime numbers and 19 total numbers.
The chances of getting a prime number both times is 7/19 of 2/5. To get the answer in one fraction, we multiply (7/19)*(2/5) = 14/95.
The answer is 14/95.
