From the 5 points A, B, C, D, and E on the number line above, 3 differ

Question

https://gmatclub.com/forum/download/file.php?id=65289 From the 5 points A, B, C, D, and E on the number line above, 3 different points are to be randomly selected. What is the probability that the coordinates of the 3 points selected will all be positive? A. \frac{1}{10} B. \frac{1}{5} C. \frac{3}{10} D. \frac{2}{5} E. \frac{3}{5} Are You Up For the Challenge: 700 Level Questions Untitled.png

BrentGMATPrepNow · Accepted Answer

Note: There are 5 points altogether. 3 are positive, and 2 are negative.

P(all 3 points are positive) = P(1st selected point is positive   AND   2nd selected point is positive   AND   3rd selected point is positive)
= P(1st selected point is positive)   x   P(2nd selected point is positive)   x   P(3rd selected point is positive)
= 3/5   x   2/4   x   1/3
= 1/10

Answer: A

uditgr · Answer

3c3/5c3 = 1/10 (3c3 = ways to select 3 positive from 3 numbers), 5c3 = (ways to select 3 numbers from 5 numbers)

NuttyIan · Answer

I did probability of selecting all 3 positive (there is only one instance of this happening) over all probabilities. Since we are selecting 3 points out of 5, it is 5C3 possibilities, or 5!/((3!)(2!)) = 10. So 1/10th.

From the 5 points A, B, C, D, and E on the number line above, 3 differ

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

From the 5 points A, B, C, D, and E on the number line above, 3 differ

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards