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 Post subject: GMAT ProbabilityPosted: Fri Jun 04, 2010 2:51 pm

Joined: Sun May 30, 2010 3:15 am
Posts: 424
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 > x²/3?

A. 1/7
B. 2/7
C. 3/7
D. 4/7
E. 1

Solution:
(C) 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 > x².

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).
-----------

Correct answer is B. 2/7
Set S = {1,2,3,4,5,6}. It doesn't include 0 as 0 is neither positive nor negative.

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 Post subject: Re: Math (test 2, question 2): probability question (normal)Posted: Fri Jun 04, 2010 2:57 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
When you take GMAT you should be very attentive to every term in the question statement. If it asked us about positive integers then the set S would be {1, 2, 3, 4, 5, 6}. But the question statement asks us about non-negative integers. Non-negative integer is an integer which is not negative. Therefore non-negative integer is either a positive integer or 0, since 0 is not negative. The set S in this case is {0, 1, 2, 3, 4, 5, 6}.

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