A community is concerned about the supply of energy for the coming winter. Suppose there are three major sources of energy for the community, namely electrical power, natural gas and oil. Let E, G, and O denote, respectively, the shortages of these sources of energy for the next winter. Also, it is estimated that the respective probabilities of these shortages are as follow:

P(E) =0.15 ; P(G)= 0.10 and P(O)= 0.20

Furthermore, if there is a shortage of oil, the probability that there will be shortage of electrical energy will be doubled. The shortage of gas may be assumed to be statistically independent of shortages in oil and electricity.

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?

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

answer: -0.5 would be the most likley outcome

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