­Each Monday through Friday (a workweek), Avinash will bring either ex

Here's how I approached this. It took me 2 mins and 28 seconds, the first full minute of which was spent deciphering what we're really being asked here. Let's start with that.

Despite the really long prompt, we're really being told a few things. First, P(Apple) = 0.5, and P(Banana) = 0.5. We're also told that Avinash correctly calculcated the probability of bringing an apple at least 4 days in a row or a banana at least 4 days in a row. Then we're asked what that probability is in the form m/n. So, we're really being asked, "what's the probability of choosing an apple 4 or 5 days in a row OR choosing a banana 4 or 5 days in a row?"

Step One:
So let's begin with bringing an apple at least 4 consecutive days in a row. In plain English, as I alluded to above, this is like bringing an apple 4 days in a row OR 5 days in a row. Using simple probabilities, we can raise (1/2)^4 to find the likelihood we bring an apple 4 days in a row and (1/2)^5 to find the likelihood we bring an apple 5 days in a row. Adding the two together (which we do for that "at least" verbiage), we get (3/32). We're halfway there. This is the probability we bring an apple at least 4 days in a row. 

Step Two:
Now we need to find the probability we bring a banana at least 4 days in a row. Here's the key, though. The probability of bringing a banana is the same as an apple. So it's the same calculation, resulting in the same probability: (3/32).

Step Three:
Combine. The prompt is asking us to find an "OR" probability, meaning we add P(Apple at least 4 days in a row) + P(Banana at least 4 days in a row): i.e. (3/32) + (3/32). 

This gives us a value of 6 for m and 32 for n.

Open to comments!­ Would hate to have gotten this right but with bad methodology.­­

mx39, you are perfectly fine with your logic. Great work.

 

Apples and Bananas are equivalent to Heads and Tails of a fair coin so we will talk in terms of Heads and Tails only. 
5 days of taking a fruit is equivalent to 5 tosses of the coin. The number of distinct outcomes will be 2^5 = 32 

P(an apple on at least 4 consecutive days) = P(Heads of at least 4 consecutive days) 
There are only 3 ways of doing this: HHHHT, THHHH, HHHHH
Note that we do not need any 4 Heads, we need 4 consecutive Heads. 
P(an apple on at least 4 consecutive days) = 3/32

Similarly, ­P(a banana on at least 4 consecutive days) = 3/32

P(an apple on at least 4 consecutive days) OR P(a banana on at least 4 consecutive days) will simply be the sum of individual probabilities 3/32 + 3/32 = 6/32
Note that there is no overlap here. Any case in which an apple was taken on at least 4 consecutive days will not allow to take a banana on 4 consecutive days. Hence "Both" will be 0 here. 

ANSWER m= 6, n = 32