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) = 0.6 and P(B) = 0.2.(a) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning.(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) 0.8

(b) 0.68

(c) 0.8824

Step-by-step explanation:

Success of Asian project, P(A) = 0.6

Success of European project, P(B) = 0.2

(a) Since both projects are independent events, the Asian project not being successful does interfere in the odds of the European project being successful. Therefore, the probability is:

(b) The probability that at least one project is successful is given by the sum of each project being successful, minus the probability of both being successful:

(c) The probability that only the Asian project is successful, given that at least one of the two projects is successful is defined as the probability of the Asian project being successful (P(A)) divided by the probability of both being successful.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}