Carry It Over I (Part 2)

by on October 2nd, 2010

Last time, I gave you this problem as an exercise in carrying information over from one mode of Quantitative thought to another.  In this particular case I combined percents and probability.

Raj is working on a set of Data Sufficiency problems for his December GMAT: a geometry problem, an algebra problem, and a data interpretation problem. He has determined that statement 1 of the geometry problem is insufficient on its own, that both statement 1 and 2 of the algebra problem are insufficient on their own, and that statement 2 of the data interpretation problem is insufficient on its own. If the probabilities are expressed as percents, approximately how much greater is the probability that all three answers are “C” after Raj figures out that statement 1 of the data interpretation problem is also insufficient on its own?

A) 2.3%
B) 2.8%
C) 3.3%
D) 5.6%
E) 8.3%

Step 1: Figure out what we need to know.  We have two probabilities that we need to figure out:  the one before Raj figures out that statement 1 of the data interpretation problem is also insufficient, and the one after.  The answer is the difference between them, in the form of a percent.

Step 2: The first probability.

  • If statement 1 of the geometry problem is insufficient, then the only valid answer choices are B, C, and E; the probability that it is C is 1/3.
  • If both statements of the algebra problem are insufficient on their own, then the only remaining valid answers are C and E; the probability that it is C is 1/2
  • If statement 2 of the data interpretation problem is insufficient, then the remaining answers are A, C, and E, and the probability that C is correct is 1/3.

The probability of all three occurring is the product of the probability fractions:  (1/3)*(1/2)*(1/3) = 1/18.

Step 3: The second probability.

Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.

The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12.

Step 4: The answer.  Note that here, as will sometimes happen on the GMAT, values that you compute “along the way” appear as wrong answer choices.  This problem calls for the difference between our two numbers, (1/12)-(1/18) = 1/36, which is a 2.77777% chance, rounded to 2.8% — the correct answer is Choice B.

Congratulations to readers Gurpreet Singh and Abdur Ashique on getting it!  I’ll try to make the next one even better.

1 comment

Ask a Question or Leave a Reply

The author Jim Jacobson gets email notifications for all questions or replies to this post.

Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a Gravatar to have your pictures show up by your comment.