Respuesta :
Answer:
0.665
Step-by-step explanation:
Given: 100 people are split into two groups 70 and 30. I group is given cough syrup treatment but second group did not.
Prob for a person to be in the cough medication group = 0.70
Out of people who received medication, 34% did not have cough
Prob for a person to be in cough medication and did not have cough
=[tex]\frac{0.70(34)}{60}=0.397[/tex]
Prob for a person to be not in cough medication and did not have cough
=[tex]\frac{0.3(20)}{30}=0.20[/tex]
Probability for a person not to have cough
= P(M1C')+P(M2C')
where M1 = event of having medication and M2 = not having medication and C' not having cough
This is because M1 and M2 are mutually exclusive and exhaustive
SO P(C') = 0.397+0.2=0.597
Hence required prob =P(M1/C') = [tex]\frac{0.397}{0.597}=0.665[/tex]
Answer:
The answer is : 63%.
Step-by-step explanation:
There are 100 people with a cough in the study.
70 patients received the cough medication, and 30 other patients did not receive treatment.
34 of the patients who received the medication reported no cough at the end of the study.
20 of the patients who did not receive medication reported no cough at the end of the study.
So, there are a total of 54 patients who reported no cough after the medication.
Out of this 54 patients, 34 took the medication.
So, the required probability is:
[tex]\frac{34}{54}\times100= 62.96[/tex]% ≈ 63%
The answer is : 63%.
