If the set S consists of all non-negative integers less than 7, what

Numbers satisfying the criteria are 0, 1, and 2, so 3/7. (C)

Set S is set of NON-NEGATIVE INTEGER less than 7 , S = {0,1,2,3,4,5,6,7}
Total Number of element in set S = 7

We need X such that 3 > x^2/3
or 9 > x^2

Only three number from set S satisfy this condition , those are 0,1,2

Hence Required Probability = Number of Element satisfying condition / Total number of element = 3/7
Hence Option C should be correct.

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).

We see that set S = {0, 1, 2, 3, 4, 5, 6}.

Let’s test each number:

3 > 0^2/3 = 0? yes

3 > 1^2/3 = 1/3? yes

3 > 2^2/3 = 4/3? yes

3 > 3^2/3 = 9/3 = 3? no

Since 3 of 7 numbers (namely, 0, 1, and 2) satisfy the inequality, the probability of getting these 3 numbers is 3/7.

Answer: C

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