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