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statuscvo [17]
3 years ago
13

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers. Roun

d to six decimal places. (Hint: these are dependent events)
Mathematics
1 answer:
AlekseyPX3 years ago
4 0

Answer:

P = 0.008908

Step-by-step explanation:

The complete question is:

The table below describes the smoking habits of a group of asthma sufferers

               Nonsmokers      Light Smoker     Heavy smoker      Total

Men                        303                     35                           37        375

Women                   413                     31                            45        489

Total                        716                     66                           82        864

If two different people are randomly selected from the 864 subjects, find the probability that they are both heavy smokers.

The number of ways in which we can select x subjects from a group of n subject is given by the combination and it is calculated as:

nCx=\frac{n!}{x!(n-x)!}

Now, there are 82C2 ways to select subjects that are both heavy smokers. Because we are going to select 2 subjects from a group of 82 heavy smokers. So, it is calculated as:

82C2=\frac{82!}{2!(82-2)!}=3321

At the same way, there are 864C2 ways to select 2 different people from the 864 subjects. It is equal to:

864C2=\frac{864!}{2!(864-2)!}=372816

Then, the probability P that two different people from the 864 subjects are both heavy smokers is:

P=\frac{82C2}{864C2}=\frac{3321}{372816}=0.008908

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Jobisdone [24]

Answer:

first off, this may be completely wrong, but I did try my best. You need to clarify what this is for so I went with the easiest route, solving for variables.

P=0

X= - 3/4

Step-by-step explanation:

P(-2)=-3+4x

Solving for X

-2P=-3+4x

-2(0)=-3+4x

-3=4x

- 3/4=X

Solving For P

P(-2)=3+4(- 3/4)

P(-2)3-3

P=0

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Answer:

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Step-by-step explanation:

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