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MrRa [10]
3 years ago
6

Bobby is testing the effectiveness of a new cough medication. There are 100 people with a cough in the study. Seventy patients r

eceived the cough medication, and 30 other patients did not receive treatment. Thirty-four of the patients who received the medication reported no cough at the end of the study. Twenty of the patients who did not receive medication reported no cough at the end of the study. What is the probability that a patient chosen at random from this study took the medication, given that they reported no cough?
Mathematics
2 answers:
Natalka [10]3 years ago
8 0

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:

\frac{34}{54}\times100= 62.96% ≈ 63%

The answer is : 63%. 

Tresset [83]3 years ago
4 0

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

=\frac{0.70(34)}{60}=0.397


Prob for a person to be not in cough medication and did not have cough

=\frac{0.3(20)}{30}=0.20

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') = \frac{0.397}{0.597}=0.665

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<h3>Further explanation</h3>

The probability of an event is defined as the possibility of an event occurring against sample space.

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<h2>Permutation ( Arrangement )</h2>

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<h2>Combination ( Selection )</h2>

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Let us tackle the problem.

This problem is similar to Pascal Triangle.

The Lord can move to the right or up in one way only.

We can put "1" in each box on the bottom and left to represent 1 possible movement.

To get the possibility of movement to another box, we will add the possibility of movement to the left, right, and diagonal squares as shown in Figure 1.

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Finally, after this process is repeated until the top right box, we get the results as shown in Figure 3.

There are 48639 ways for a lord to move from the bottom left to top right corner of the 8 by 8 chessboard.

<h3>Learn more</h3>
  • Different Birthdays : brainly.com/question/7567074
  • Dependent or Independent Events : brainly.com/question/12029535
  • Mutually exclusive : brainly.com/question/3464581

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Probability

Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation

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