For each of four symptoms, the graph shows the percentage chances that

chetan2u Sajjad1994 Bunuel
how are we going to calculate the probability here?
Please post explanation for this

But for the case of Day 2 and Day 6, only sum of "nasal drainage" and "sore throat" can be greater than 0.98, so why "exactly two" can't meet the answer?

ps:
on Day 10, the sum of percentage (30% and around 20%) is only 0.5 that is less than 0.98. Why Day is the the right answer?

litingyxsh did you ge the answer

chetan2u plz post the official solution­

KarishmaB chetan2u Bunuel

For Day-2:
Nasal drainage = 60%
Sore throat = almost 50%
Cough = 38%
Fever = 15 %

If i have to calculate ''one or fewer'' for Day-2, how do i do it? Is the below process correct?

One of fewer = [1 - (ONLY 2 + ONLY 3 + ONLY 4 )symptoms]
Only TWO: ((0.6 x .50) + (0.60 x 38) +(0.60 x 0.15) + (0.50 x 0.38) + (0.50 x 0.15) +(0.38 x 0.15))
Only THREE: Likewise taken 3 at a time
Only Four: (0.60 x 0.50 x 0.38 x 0.15)­

Moreover, how would you attempt this question?

 

­Only TWO equation is not correct. 
((0.6 x .50) + (0.60 x 38) +(0.60 x 0.15) + (0.50 x 0.38) + (0.50 x 0.15) +(0.38 x 0.15))
While writing 0.6. x .50,
If I am right, you are considering cases when there is ONLY nasal & sore throat. Instead of 0.6 x 0.5 , it should be 0.6 x .5 x .62 x .85 
ONLY nasal & sore throat = (have nasal) x (have sore throat) x (DON'T HAVE cough) X (DON'T HAVE fever)

Meanwhile I will not approach this question this way as it would be too much calculation. I have given solution above with id "manasp35". Let me know if you have any doubts.  

­Hi manasp35
Thanks for addressing. This actually helps. 
Infact, i tried to solve the above case using your equation in the previous poster. I couldn't have thought along those lines. Good work.

For me if you understand the question correctly you don't have to put too much thought.

First understanding that over 0.98 represents 1 in reality because less than 1% don't show clears the field. ( from this if you are observant enough you would assume that the answer falls on day 10 cause it lacks 2 symptoms)

After that you should try finding a scenario that forces 1.

Intuitively from the choices 1 or more and 2 or less would be the first tries cause you have the wider range of probable scenarios each day. However, 1 or more is easily excluded without much calculation and day 10, 2 or less symptoms is easy to catch.

Official answer:
On Day 9, the lines for fever and sore throat stop. This means that, for each of these symptoms, there is less than a 1% chance that someone with a cold will have the symptom beyond Day 9. Therefore, on Day 10, the chance, p, that someone with a cold will have at least one of these symptoms is less than 2%, and they will have at most two (that is, two or fewer of the remaining symptoms). Also, from p < 0.02, it follows that 1 − p > 0.98. Thus, the probability is greater than 0.98 that someone with a cold will have two or fewer of the four symptoms on Day 10.

We need higher than 98% probability of something. That is as good as certainty, including certainty. 

What am I almost certain about? Am I certain that on Day 2, a person will have at least 1 symptom? No. On Day 2, there is 62% probability of nasal drainage, 49% of sore throat, 38% of cough and 16% of fever. What if all other probabilities lie within the circle of nasal drainage? That is the same 62% people see other symptoms also? Then the probability of a person showing at least one symptom will be 62% only. 

But what I am certain about is that at Day 10, a person can show at most 2 symptoms. For the other 2 symptoms, the probability is less than 1%. Even if I assume it to be almost 1% and different set of people show the third and fourth symptoms, still there is a more than 98% probability that a person shows 2 or fewer symptoms on Day 10. 

ANSWER: two or fewer symptoms on Day 10

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