An oil exploration company currently has two active projects, one in asia and the other in europe. let a be the event that the asian project is successful and b be the event that the european project is successful. suppose that a and b are independent events with p(a) = 0.2 and p(b) = 0.5.

(a) if the asian project is not successful, what is the probability that the european project is also not successful?



explain your reasoning.
a. since the events are independent, then a' and b' are independent, too.
b. since the events are independent, then a' and b' are mutually exclusive.
c. since the events are not independent, then a' and b' are mutually exclusive.
d. since the events are independent, then a' and b' are not independent.

(b) what is the probability that at least one of the two projects will be successful?

(c) given that at least one of the two projects is successful, what is the probability that only the asian project is successful?

a)

0.5

option A

b)

0.6

c)

0.1

Step-by-step explanation:

The event A and B are independent so

P(A∩B)=P(A)*P(B)

P(A∩B)=P(0.2)*P(0.5)=0.10

a)

We have to find P(B'|A')

P(B'|A')=P(B'∩A')/P(A')

P(A)=0.2

P(A')=Asian project is not successful=1-P(A)=1-0.2=0.8

P(B)=0.5

P(B')=Europe project is not successful=1-P(B)=1-0.5=0.5

P(B'∩A')=Europe and Asia both project are not successful=P(A')*P(B')=0.8*0.5=0.4

P(B'|A')=P(B'∩A')/P(A')=0.4/0.8=0.5

This can be done by another independence property for conditional probability

P(B|A)=P(B)

P(B'|A')=P(B')

P(B'|A')=0.5

b)

Probability of at least one of two  projects will be successful means that the probability of success of Asia project or probability of success of Europe project  or probability of success of Europe and Asian project which is P(AUB).

P(AUB)=P(A)+P(B)-P(A∩B)

P(AUB)=0.2+0.5-0.1

P(AUB)=0.6

c)

Probability of only Asian project is successful given that at least one of the two projects is successful means that probability of success of project Asia while the project Europe is not successful denoted as P((A∩B')/(A∪B))=?

P((A∩B')/(A∪B))=P((A∩B')∩(A∪B))/P(A∪B)

P((A∩B')∩(A∪B))=P(A∩B')*P(A∪B)

P(A∩B')=P(A)*P(B')=0.2*0.5=0.10

P((A∩B')∩(A∪B))=0.1*0.6=0.06

P((A∩B')/(A∪B))=0.06/0.6=0.1

the answer to this question is yes

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step-by-step explanation: