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

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?

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.

­It can be determined from the graph that the probability is greater than 0.98 that someone with a cold will have [two or fewer] of the four symptoms on day [10]

On day 10, most show at most 2 symptoms (nasal drainage and sore throat)

On day 10, the probability for each of 2 symptoms cough and fever is < 1%

Even the group having these symptoms (cough and fever) are different set, that adds up to 2% 

=> 98% will have at most 2 symptoms (nasal drainage and sore throat)
 


 

­KarishmaB can you explain why "Only Two symptoms" is incorrect for Day 10?

On Day 10, there is about 20% chance of nasal drainage and about 30% chance of cough. ­So a person would have at most 2 symptoms. He can have 1 or 0 symptoms too. Hence 'Exactly two' is not correct. After all, there is only 20% chance of nasal drainage and only 30% chance of having cough on day10. 

We need day 10 because for we need sum of p*(1-p)... to be almost close to 1, 
as we have p<0.01 for 2 cases , 1-p will be >0.99 for sore and fever.

Hence for day 10:

we have P1<0.01, P2<0.01, P3=0.2, P4=0.3

For 2 or fewer : we get max coverage of (1-p) in sum of probabilities:
we have (1-p1)*(1-p2)+...
this will get us the max probabilty.
calculation of this is very lengthy.

0.99*0.2+0.99*0.3+0.99*0.99+0.99*0.8+...

Hence we can say Day 10 & two or fewer­

­Hi KarishmaB
I am unable to understand this part- 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.
Can you please help me understand how we are extending the possibility of someone showing nasal drainage as a symptom to 62% of the people are suffering from nasal drainage?

I selected exactly two. But after reviewing the solution, I think it is wrong because of the following reason-
Like you mentioned, the graph talks about likelihood/certainty. On day 10, there is less than 2% probability that sore throat and fever will persist. But I cannot say that the probability is 98% for EXACTLY 2 symptoms. Say nasal drainage is event A and cough is event B, 98% = A u B. Out of this I cannot say what is the probability of exactly 1. Like you mentioned, all nasal drainage cases may be covered under the cough circle or not.
KarishmaB chetan2u MartyMurray can you please let me know if my thought process is correct here?

dear KarishmaB, @chetan2u


I am afraid I did get what the graph says.

say on day 10, the chance of getting sore throat is 21%, that of getting cough is 29%
IMO , the chance of getting either sore throat or cough or both is most 29%



ANSWER: two or fewer symptoms on Day 10, which confused me a lot

­would you experts explain further ?

have a nice day

On Day 10

chance of getting -

nasal drainage is 21%
cough is 29%
sore throat is about 0% (less than 1)
fever is about 0% (less than 1)

What is the probability of getting 3 or more of the given symptoms? Since chance of both sore throat and fever is about 0, chance of getting 3 or all 4 symptoms is about 0.
This means that chance of getting 2 or 1 or 0 symptoms is almost 100% i.e. more than 98%. So almost certainly one would have either 0 or 1 or at most 2 symptoms. The chance of a third symptom is almost 0 so having 3 or more symptoms will be very rare.

Hence the answer is chance of "2 or fewer" is more than 98%.

Required: Probability for someone WITH cold >= 0.98
element = Number of symptoms.
Hence, Probability ("Num of symptoms" being two or less) on day 10 = almost 100%.
Don't get confused by extra data. It is NOT probability of cold BUT probability of number of symptoms.

This would be true only if we know that the symptoms occuring are independent events which is not given in the question.

So we may not be able to calculate ONLY TWO , ONLY THREE, ONLY FOUR for this question with the information provided. We can try to find the maximum probability of exactly 2 events occuring but for that this question will become a Linear Programming Problem ( which is not required / out of scope I would say )

BEST WAY TO SOLVE - If we think about the probabilities as venn diagrams , then we can side step thinking about wheather the events are dependent or independent. We just need to think about minimising overlaps and maximising overlaps , and in this way we can get ranges for any probability and eliminate wrong options / indetify the correct option.

More than the math I am baffled by the language in these questions. lol
these are becoming less math and more English

For making it easier, I simplified some numbers like 98% -> 100%.

<Day 10>
symptom A - 30%
symptom B - 20%
symptom C - 0%
symptom D -0%

-> Someone with a cold will have 3 or more of the four symptoms on day 10 ­is 0%.

That means someone with a cold will have 2 or fewer of the four symptoms on day 10 is 100%