英语数学概率题帮同学问的,if the probability of successfully introducing
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英语数学概率题
帮同学问的,
if the probability of successfully introducing any new product(A for example) to the market is 18%,and the probability of successfully introducing two new products(A and B) is 5%,then
a) what is the probability that any new product (A for example) will fail?
b) what is the probability that at least one of the two( any two),A or B,will be successful
c) what is the probability that neither A nor B will be successful
d) what is the probability that A will be successful,if you know that B was successful
e) waht is the probability that A or B was successful,but not both
f) Given that at least one,A or B,was successful,what is the probability that A was successful
帮同学问的,
if the probability of successfully introducing any new product(A for example) to the market is 18%,and the probability of successfully introducing two new products(A and B) is 5%,then
a) what is the probability that any new product (A for example) will fail?
b) what is the probability that at least one of the two( any two),A or B,will be successful
c) what is the probability that neither A nor B will be successful
d) what is the probability that A will be successful,if you know that B was successful
e) waht is the probability that A or B was successful,but not both
f) Given that at least one,A or B,was successful,what is the probability that A was successful
Acording to P(A)=18% P(AB)=5%,we come to the conclusion that
P(A')=1-P(A)=82% and P(AB')=P(A)-P(AB)=18%-5%=13%.
Similarly,P(B)=18% and P(BA')=13%.
Then,P(A'B')=1-P(AB')-P(BA')-P(AB)=1-13%-13%-5%=69%
answers to these questions:
a)P(A')=1-P(A)=82%
b)1-P(A'B')=31%
c)P(A'B')=69%
d)on the basis of Conditional Probability Formula:
P(AB)/P(B)=5%/13%=5/13≈38.5%
e)P(AB')+P(BA')=13%+13%=26%
f)It is similar to d):
P(A)/[1-P(A'B')]=5%/31%=5/31≈16.1%
End!
HaHa,my chinglish looks amateur!
P(A')=1-P(A)=82% and P(AB')=P(A)-P(AB)=18%-5%=13%.
Similarly,P(B)=18% and P(BA')=13%.
Then,P(A'B')=1-P(AB')-P(BA')-P(AB)=1-13%-13%-5%=69%
answers to these questions:
a)P(A')=1-P(A)=82%
b)1-P(A'B')=31%
c)P(A'B')=69%
d)on the basis of Conditional Probability Formula:
P(AB)/P(B)=5%/13%=5/13≈38.5%
e)P(AB')+P(BA')=13%+13%=26%
f)It is similar to d):
P(A)/[1-P(A'B')]=5%/31%=5/31≈16.1%
End!
HaHa,my chinglish looks amateur!
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