Manhattan GMAT Challenge Problem of the Week – 19 November 2012

by on November 19th, 2012

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5/6 of the population of the country of Venezia lives in Montague Province, while the rest lives in Capulet Province. In the upcoming election, 80% of Montague residents support Romeo, while 70% of Capulet residents support Juliet; each resident of Venezia supports exactly one of these two candidates. Rounded if necessary to the nearest percent, the probability that a Juliet supporter chosen at random resides in Capulet is

A. 28%
B. 41%
C. 45%
D. 72%
E. 78%


Pick 60 as the “smart number” for the total population of Venezia. Then 50 (= 5/6 of 60) people reside in Montague, and the other 10 reside in Capulet. Now compute the number of Juliet supporters in each province. Since 80% of Montague residents support Romeo, the other 20% must support Juliet (remember, 100% support one of these two candidates):

10 Montague residents support Juliet.

We can compute the number of Capulet residents that support Juliet directly:

70% of 10 Capulet residents = 7 that support Juliet.

Thus, there are a total of 10 + 7 = 17 Juliet supporters.

Picking at random from among these 17, we would only get a resident of Capulet 7/17 of the time. Perform the long division and round to 41%.

Notice that although Montague residents are much more likely to support Romeo, there are so many more residents of Montague that the Juliet supporters among them outnumber the Juliet supporters in her “home base” of Capulet. However, knowing that the person we’re picking is a Juliet supporter raises the probability of that he or she resides in Capulet—after all, without knowing who is supported, we’d give a 1/6 = ~17% chance of picking a resident of Capulet. After knowing that the person supports Juliet, we increase the chance to 41%.

By the way, the math you’ve done here has the fancy name of Bayes’ Theorem, which you’ll encounter again in business school in your Statistics class.

The correct answer is B.

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