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

Question

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For each of four symptoms, the graph shows the percentage chances that someone will have that symptom on each day of the 2 weeks after first developing a cold. The lines for certain symptoms do not continue past a certain day because the chance that someone will have those symptoms beyond that day is less than 1%.

Select the options from the drop-down menus that create the statement that is most strongly supported by the information provided.

It can be determined from the graph that the probability is greater than 0.98 that someone with a cold will have  GRAPH_1_PLACEHOLDER  of the four symptoms on day  GRAPH_2_PLACEHOLDER .

ID: 700216

manasp35 · Accepted Answer

Total probability = P(no symp) + P (exactly 1) + P (exactly 2) + P(exactly 3) + P (exactly 4)

We need to find probability that is greater than 0.98 , meaning almost 1. This implies that almost covers all the cases.

Acc. to options, we have Day 2 , Day 6 , Day 10

For Day 2 and Day 6, 
Total probability = P(no symp) + P (exactly 1) + P (exactly 2) + P(exactly 3) + P (exactly 4)

Options given: one or fewer, one or more, two or fewer, exactly two

None of the options cover all the cases.

For Day 10,
Total probability = P(no symp) + P (exactly 1) + P (exactly 2)

In this case, two or fewer cover all the cases.

answer: two or fewer, Day 10

KarishmaB · Answer

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

litingyxsh · Answer

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?

ashutosh_73 · 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 =  
 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?

manasp35 · Answer

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.

ashutosh_73 · Answer

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.

thecardinal · Answer

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.

ruis · Answer

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.

Gemmie · Answer

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

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)

Freddy12 · Answer

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

KarishmaB · Answer

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.

Oppenheimer1945 · Answer

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

RenB · Answer

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?

zoezhuyan · Answer

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

KarishmaB · Answer

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%.

SmileAndSolve · Answer

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.

UDEMYtutorJackson · Answer

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.

D3N0 · Answer

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

jin.h · Answer

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%

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

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