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Sauron [17]
3 years ago
10

According to a survey done at your school, about 42% of all the female students participate in 2 sport seasons. You randomly ask

5 female students at your school how many sport seasons they participate in. What is the probability that at least 3 of the 5 female students surveyed participate in 2 sport seasons? Show your work.
If anyone can answer this in the next 30 minutes, please do!!
Mathematics
1 answer:
Leni [432]3 years ago
5 0

Answer:

Hi Christine

Step-by-step explanation:

Ms. Paik would be mad, wouldn't she?

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe
nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let D^c be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that  one person in 10,000 people has a rare genetic disease.

So,P(D)=\frac{1}{10000}

Only 0.4% of the people who don't have it test positive.

P(A|D^c) = probability that a person is tested positive given he don't have a disease = 0.004

P(D^c)=1-\frac{1}{10000}

Formula:P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}

P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}

P(D|A)=\frac{2470}{2471}=0.9995

P(D|A)=0.9995

A)The probability that someone who tests positive has the disease is 0.9995

(B)

P(D^c|A^c)=probability that someone does not have disease given that he tests negative

P(A^c|D^c)=probability that a person tests negative given that he does not have disease =1-0.004

=0.996

P(A^c|D)=probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}

P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}

P(D^c|A^c)=0.99999

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0
3 years ago
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