Help me with this probability question, please?

Question

Help me with this question, please?
And explain/show how to get the result? :/
Thank you a bunch. <3

What is the probability that the last three digits of a randomly selected phone number are all prime? Express your answer as a common fraction.

Thanks.

Answer 1

Hi,

8/125 is the probability that the last three digits of a randomly selected phone number are all prime.

This is because the prime digits are 2, 3, 5, and 7. That means there are 4 primes out of the 10 digits. So, in each of the 3 positions there is a 4 out of 10 chance of a prime, which is the same as a 2/5 chance for each digit. For all 3 digits, the probability of all primes is:

2/5 x 2/5 x 2/5 = 8/125

I hope that helps!! :-)

Answer 2

We assume that the numbers 0-9 appear randomly in these 3 digits. The prime digits are 1,2,3,5,and 7. So p(any digit prime) = 1/2 By the above assumption, p(three in a row prime)= 1/8

Answer 3

4 of the 10 digits are prime -- 2, 3, 5, 7 -- so the prob a digit is prime is 4/10 = 2/5. prob last 3 prime is (2/5)^3 = 8/125, assuming it's OK that they not be different.