# Manhattan GMAT Challenge Problem of the Week – 5 Aug 2010

by on August 5th, 2010

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.

As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

## Question

Every trading day, the price of CF Corp stock either goes up by \$1 or goes down by \$1 with equal likelihood. At the end of 5 trading days, what is the probability that the price of CF Corp stock is up by exactly \$3 from its initial price?

A. 1/16
B. 1/8
C. 5/32
D. 9/32
E. 3/8

The daily change in CF Corp’s stock price can be compared to a coin flip. Heads – the price goes up by \$1. Tails – the price goes down by \$1. Moreover, the coin is “fair”: that is, each daily outcome is equally possible (meaning that the chance of heads is 50%, and the chance of tails is 50%). This also means that any particular sequence of flips is equally probable. As a result, our probability calculation is simplified. We just count the 5-day sequences that give us a \$3 increase, then we compare that number to the total number of 5-day sequences.

To go up exactly \$3, we need exactly 4 “up” days (heads) and 1 “down” day (tails). We don’t care about the order in which these days come. So we need to count the possible arrangements of 4 heads and 1 tails. We can simply list these out:

HHHHT

HHHTH

HHTHH

HTHHH

THHHH

The only question is what day of the week the “down” day falls on, so there are 5 possibilities for a \$3 increase. Alternatively, we can use the combinations or anagrams method to calculate (5!)/(4!) = 5.

Now, we need to count all the possible 5-day sequences of flips. Since each day can have 2 outcomes (H or T), we have 2 × 2 × 2 × 2 × 2 = 32 total possible outcomes over 5 days.

Finally, to compute the probability of a \$3 increase over 5 days, we divide 5 successful outcomes by 32 total possibilities (of equal weight) to get 5/32.