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805+ Level|   Math Related|            
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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.­­
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mx39, you are perfectly fine with your logic. Great work.
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­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.

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
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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.
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­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: \(\frac{6}{32} = \frac{m}{n}\)­
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Keeping it short, with more mathematical denotions and less Reading comprehension :lol::

Apple=HHHHT + HHHHH + THHHH=3/2^5

Banana=TTTTH + TTTTT + HTTTT=3/2^5

P=6/32
m=6, n=32­
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Quote:
he would bring whatever on at least 4 consecutive days
\(p=5×0.5^4×0.5^1+1×0.5^5×0.5^0=6×0.5^5=\frac{6}{2^5}=\frac{6}{32}\)­
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Hi, just wondering why this can't be:

[Prob to find exactly 4 days]
5C4*(1/2)^4*(1/2)
<either apple or banana>
2*5C4*(1/2)^4*(1/2)

[Prob to find exactly 5 days]
5C5 * (1/2)^5
<either apple or banana>
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).
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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

Axwe7
Hi, just wondering why this can't be:

[Prob to find exactly 4 days]
5C4*(1/2)^4*(1/2)
<either apple or banana>
2*5C4*(1/2)^4*(1/2)

[Prob to find exactly 5 days]
5C5 * (1/2)^5
<either apple or banana>
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).
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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?
mx39
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.­
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(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.

movehs
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?
mx39
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.­
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