An emergency room contains 7 sickly children, 3 healthy children, 7 sickly adult

Question

An emergency room contains 7 sickly children, 3 healthy children, 7 sickly adults, 3 healthy adults, and 5 sickly elderly people. If we randomly select one person from this sample, find the following probabilities: i) P(child) ii) P(sickly) iii) P(healthy and elderly) iv) P(sickly or adult) v.) P(sickly)

If we use the emergency room from question 1, calculate the following:

How many ways can we sample 10 people if order is not important?

How many ways can sample 5 people if order is important?

How many different ways can we sample 3 people if we replace each person after sampling him/her?

Explanation / Answer

total sample space =7+3+7+3+5 =25

i)P(child) =( sickly children+ healthy children)/total people =(7+3)/25 =10/25 =0.4

ii)P(sickly) =(sickly children+sickly adult+sickly elderly people)/total people =(7+7+5)/25=19/25=0.76

iii)

P(healthy and elderly) =0 (as there is 0 person who are elderly and healthy)

iv)

P(sickly or adult) =P(sickly)+P(adult)-P(sickly and adult)=(7+7+5)/25+(7+3)/25-(7/25)=0.88

v)

P(sickly) =(7+7+5)/25=0.76

2) number of ways to sample 10 people =25C10 =25!/(10!*(25-10)!) =3268760

3)

number of ways to sample 5 if order is important =25P5 =25!/(25-5)! =6375600

4)

number of ways to sample =25*25*25 =15625 (As for each sampling unit 25 choices are there)

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