mehrasa wrote:1) A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?
1/4 1/2 3/4 15/16 1/16

There are 8 digits to choose from (0, 2, 3, 5, 6, 7, 8 and 9), 4 of which are odd. Assuming we pick our codes randomly, and that we can use repeated digits, the probability any individual digit is odd is 4/8 = 1/2. The probability all are odd is thus just (1/2)(1/2)(1/2)(1/2) = 1/16, and the probability at least one digit is even is then 1 - 1/16 = 15/16.

mehrasa wrote: 2)John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?
15/16 11/16 11/12 1/2 5/8

[spoiler]OA: B[/spoiler]

The wording of this question is terrible. When it says "the digit 1 appeared in the last three places", this could mean one of two things: that *all* of the last three digits are '1', or that *one* of the last three digits is '1'. The former meaning is what's intended. You are also meant to assume that none of the other digits is '1', something which is completely unclear from the wording of the question. It's one of those practice questions that's more confusing than helpful. In any case, if you have psychic abilities and can work out what the question designer was thinking, you can solve it, as was done in this thread:

https://www.beatthegmat.com/probability-t17876.html