Bunuel
If the set S consists of all non-negative integers less than 7, what is the probability that a randomly selected integer x of set S will satisfy the inequality 3 > x^2/3?
A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1
Kudos for a correct solution. 800score Official Solution:We are told that S = {0, 1, 2, 3, 4, 5, 6}, that is, all integers from 0 to 6. There are 7 integers in the set. Remember that zero is an integer that is neither negative nor positive.
Now, the expression can be simplified by multiplying both sides by 3. We can do this since the 3 in the denominator is a positive number.
Remember: When you multiply or divide both sides of an inequality by a negative number, you must change the direction of the inequality.
The expression is simplified to 9 > x^2.
The only numbers in S that can be squared and satisfy this condition are 0, 1, and 2. Therefore, 3 of the 7 integers in the set will satisfy the condition, and the desired probability is 3/7.
The correct answer is choice (C).