SOLUTION: b. Hospital records show that 75% of the patients will recover from a certain disease. i.What is the probability that 2 out of 6 patients selected at random will die from a certa

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Question 1197316: b. Hospital records show that 75% of the patients will recover from a certain disease.
i.What is the probability that 2 out of 6 patients selected at random will die from a certain disease?
ii. What is the probability that 4 out of the 5 patients selected at random will recover from a certain disease?
c. Five of 10 new buildings in a city violate the building code. A random sample of three buildings
selected for inspection.
*)What is the probability of none of the buildings violate the building code in a sample of 3 buildings

*)What is the probability of at least one of the new buildings violate the building code in a sample of 3 buildings?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi  
Binomial Probability:  n = 6, p(recover) = .75, p(die) = .25
what then is the probability that the inspector's sample contains:
Using TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(2 die) =.2966 (calculator)
0r
P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+ 

 P(2die) = 15%28.25%5E2%29%28.75%5E4%29 =.2966
..............
n = 5, p(recover) = .75, p(die) = .25
P(x = 4 recover ) = .3955(calculator)   Or ++5%28.75%5E4%29%28.25%5E1%29%29 = .3955
...............
n = 3, p(violate code) = 5/10 = .5
P(x=0) = .125 (calculator)
P(at least one violate) = 1 - P(x=0) = 1 - .125 = .875
Wish You the Best in your Studies.

Wish You the Best in your Studies.