Refer to the figure at the right. Line a is
parallel to line b and mZ2 is 135º. Find
each given angle measure. Justify your
answer. (Examples 1, 2, and 4)
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a) P ( E & G & O ) = 0.006

b) P ( E U G ) = 0.235

c) P ( G&O | E ) =  0.04  

d)  P(At-least 2) = 0.101

Step-by-step explanation:

Given:

- P(E) = 0.15

- P(G) = 0.10                 independent from both E and O

- P(O) = 0.20      

- P(E/O) = 0.30

Find:

a. What is the probability that there will be shortages in all the three sources of energy next winter?

b. What is the probability that there will be shortages in at least one of the following sources next winter: gas and  electricity?

c. If there is a shortage of electricity next winter, what is the probability that there will also be shortages in both gas and Oil?

d. What is the probability that at least two of the three sources of energy will be in short supply next winter?

Solution:

- The probability requested is P ( E & G & O ).

  Use the conditional probability given P(E/O) to formulate P(E & O)

                           P(E&O) = P(E/O) * P(O)

                           P(E&O) = 0.30 * 0.20 = 0.06

  We known that G is independent from both E and O. Hence, using definition of independent events we can compute P ( E & G & O ):

                           P ( E & G & O ) = P(E&O) * P(G)

                           P ( E & G & O ) = 0.06 * 0.1 = 0.006

- The probability requested is P ( E U G ).

                            P ( E U G ) = P(E) + P(G) - P(E&G)

Where,                 P(E&G) = P(E)*P(G)       independent events

                            P ( E U G ) = P(E) + P(G) - P(E)*P(G)

Input the probs:

                           P ( E U G ) = 0.15 + 0.1 - 0.15*0.1

                           P ( E U G ) = 0.235

- The probability requested is P ( G&O | E ).

 Using conditional probability we have:

                            P ( G&O | E ) =  P ( E & G & O ) / P(E)

                            P ( G&O | E ) =  0.006 / 0.15

                            P ( G&O | E ) =  0.04  

- The probability requested is

                                    P(At-least 2) = P( = 2) + P(= 3)

          P(At-least 2) = P ( G & O ) + P ( G & E ) + P (O&E) + P ( E & G & O )

                                = 0.1*0.2 + 0.1*0.15 + 0.06 + 0.006

          P(At-least 2)  = 0.101