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

Question

Each Monday through Friday (a  workweek ), Avinash will bring either exactly one apple or exactly one banana with him to his workplace for an afternoon snack. To avoid having to decide which to bring, each morning Avinash will toss a coin with a face on exactly one side that is equally likely to land faceup or facedown. If the coin lands faceup, then he will bring an apple, and if the coin lands facedown, then he will bring a banana. Avinash correctly determined the probability that, for a given workweek, either he would bring an apple on at least 4 consecutive days or he would bring a banana on at least 4 consecutive days. This probability was  m  divided by  n .Select for  m  and for  n  values jointly consistent with the given information. Make only two selections, one in each column.

ID: 700259

egmat · Accepted Answer

In simple words – this is a Quant Probability question in the garb of a TPA question. So, as long as you have solid quant skills, you should be able to solve this question with utmost ease.

So, if you falter in this question, you should implement the stated corrective actions and once you do that not only will your ability in TPA improve but also in quant Arithmetic. 😊
 
 If you were not able to translate the scenario into probability equation, then the issue is your lack of conceptual understanding related to probability.
 
 Corrective Action: Build your probability conceptual understanding.   
 If you were not able to arrive at the equation that the required probability is the sum of three probabilities as stated, then the issue is that you did not consider all cases.
 
 Corrective Action: Slow down and understand the scenario presented – at least 4 consecutive days means 4 and 5 days.   
 If you did not take the word “consecutive” into account, then the issue is that you did not take into account the constraint established by the scenario.
 
 Corrective Action: Slow down while reading the scenario and put yourself in the scenario. That always helps with ‘owning the dataset’.

Here is the video solution for you!

https://www.youtube.com/watch?v=2eRCq8Au8C4

chetan2u · Answer

Another difficult level question.
Easy, if understood well.

probability as  \frac{m}{n}  means favourable outcomes divided by total outcomes.

We can concentrate on just the coin as the banana and apple depend on the outcome of throw of the coin.
Now, the coin will show exactly one of the two faces, so total ways would  be 2*2*2*2*2 or  2^5 .
Each 2 means Head or Tail, and five times multiplication corresponds to give days.

FOUR consecutive heads could be first to fourth day or second to fifth day. FIVE consecutive heads would be be first to fifth day.
Thus, 3 ways. 
Similarly, 3 ways for tails. 
Thus, favourable outcomes are 3+3 or 6.

P=  \frac{6}{32} , that is m = 6 and n = 32

mx39 · Answer

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.

chetan2u · Answer

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

KarishmaB · Answer

Video Solution to this question: https://youtu.be/sOzRtWGu3G0 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

playthegame · Answer

Thanks I ended up calculating as 12/32. I counted the days which are not consecutive.

Meaning instead of 3/32 I calculated 6/32 times 2 which is 12/32. So I chose 4 out of 5 days plus 5 out of 5 days. So 5C4 is 6 which is where I ended up counting the days that are not consecutive.

Better luck next time to me.

Gemmie · Answer

There are two possibilities for each day: apple or banana.
So, for a workweek (5 days), there are a total of  2^5 = 32  possible scenarios (combinations of apple/banana for each day).
=> n = 32

However, we only care about scenarios where Avinash has
(i) at least 4 consecutive apples OR
(ii) at least 4 consecutive bananas.

Count these favorable scenarios (m):

(1) Exactly 4 consecutive apples: There are 2 places for this streak to begin (Monday and Tuesđay). 
=> 2 favorable scenarios.

(2) Exactly 4 consecutive bananas: Similar to apples, there are 2 places for this streak to begin
=> 2 favorable scenarios.

(3) 5 consecutive apples => 1 favorable scenario

(4) 5 consecutive oranges => 1 favorable scenario

=> Probability:  \frac{6}{32} = \frac{m}{n}

Oppenheimer1945 · Answer

Keeping it short, with more mathematical denotions and less Reading comprehension

: Apple=HHHHT + HHHHH + THHHH=3/2^5 Banana=TTTTH + TTTTT + HTTTT=3/2^5 P=6/32 m=6, n=32

nisen20 · Answer

p=5×0.5^4×0.5^1+1×0.5^5×0.5^0=6×0.5^5=\frac{6}{2^5}=\frac{6}{32}

Axwe7 · Answer

Hi, just wondering why this can't be: 5C4*(1/2)^4*(1/2) 2*5C4*(1/2)^4*(1/2) 5C5 * (1/2)^5 2*5C5*(1/2)^5 Total answer is: 2*5C4*(1/2)^4*(1/2) + 2*5C5*(1/2)^5 = 6/16 (rather than answer of 6/32).

ManifestDreamMBA · Answer

The key is "atleast  consecutive days ", which is different from just picking 4 days. With 4 consecutive days, there are only 2 options instead of 5 for each case (banana or apple)

Total favourable cases (at least 4 consecutive days) are AAAAB, BAAAA, AAAAA for apple and similarly BBBBA, ABBBB, BBBBB for banana, which is equal to 6
And denominator is 2^5 = 32
P = 6/32

movehs · Answer

How come you don't multiply (1/2)^4 by 2 to account for bringing on the last 4 days of the workweek or bringing on the first four days of the workweek?

cheshire · Answer

(1/2)4×2=Probability of 4 consecutive apples (or bananas) on Mon–Thu or Tue–Fri only
You're only capturing these two cases:
 
  AAAAx  (first four days are apples) 
  xAAAA  (last four days are apples)  
This ignores:
 
  AAAAA  (which satisfies both, but is only counted once in your calculation) 
 Any case of  bananas  (BBBBx, xBBBB, or BBBBB) 
 Other patterns with 4 in the  middle  (like AAAAA)  
So this method  misses valid outcomes , and  assumes no overlap or mutual cases , which is not true.

MANASH94 · Answer

How I tried this question was: 
total possibilities is : 2^5 =32 (because 5 work days.)
Now, what are the possible combinations: 
(AAAAB, BAAAA, AAAAA, BBBBA, ABBBB, BBBBB)
So 6 total possibilities is what I counted.
Hence probability should be 6 / 32.
M = 6,
N = 32.

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

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