Source: MGMAT Challenge 10/31/05

Can you please explain what I'm doing wrong in my solution below? I understand where the problem is and I also understand the logic as per the official solution but I do not understand why the way I calculated the arrangements (using the anagram method) is wrong. Can you please elaborate? In what type of a question will I use my method? Thanks a lot.


In October 2005, there was a record 75% chance of rain in New York City on any given day (along with a 25% chance of no rain). Assuming that each day’s weather was independent of the weather on any other day, what was the probability that in a given week, it rained at least 5 days in a row?

My solution:

p(rain) = 3/4
p(sunny) = 1/4

We have 3 anagrams for 3 scenarios: 5 days of rain, 6 days of rain and 7 days of rain. (R= rain, S = sunny)

Type I:

M T W Th F S Su
R R R R R S S

= 7!/5!*2! * (3/4)^5 * (1/4)^2

Type II:

M T W Th F S Su
R R R R R R S

= 7!/6! * (3/4)^6 * 1/4

Type III:

M T W Th F S Su
R R R R R R R

= 7!/7! * (3/4)^7

total probability = 7!/5!*2! * (3/4)^5 * (1/4)^2 + 7!/6! * (3/4)^6 * 1/4 + 1 * (3/4)^7

Actual Solution:

There are 3 different types of scenarios that we need to consider: (1) 5R’s 2S’s (2) 6R’s 1S and (3) 7S’s. Of course we must remember that within each scenario it must rain for at least 5 consecutive days.

TYPE 1: There are 3 ways for it to rain exactly 5 out of 7 days and for the 5 days to be consecutive: RRRRRSS, SSRRRRR, SRRRRRS (two S’s at the end, the beginning, or one on each end).

TYPE 2: There are 4 ways for it to rain exactly 6 out of 7 days and for at least five of the rainy days to be consecutive: RRRRRRS, SRRRRRR, RSRRRRR, RRRRRSR (S in position 1, 2, 6 or 7).

TYPE 3: There is one way for it to rain exactly 7 out of 7 days: RRRRRRR.

total probability = 3 * (3/4)^5 * (1/4)^2 + 4* (3/4)^6 * 1/4 + 1 * (3/4)^7

Looking at your Type 1, you use 7!/(5!2!) - the problem with this is that it does not ensure 5 consecutive days of rain, only 5 days of rain within the 7 days. So you're counting RRRRRSS (which should be counted, according to the problem), but you're also counting RSRRRSR (which should not be counted, b/c it doesn't have 5 consecutive R's).

Generally, when you have the kinds of bizarre constraints that you see in this problem, formulas (or the anagram method) will get you into trouble. Better to logic it out. But, honestly - you won't see something like this on the test, so don't worry about it. A lot of the challenge problems are harder than anything on the real test and weird combinatorics questions are just not that common.

Thanks a lot for the explanation Stacey; it is very helpful.

You're welcome!