An explosion in a liquefied natural gas tank undergoing repair could have occurred as the result of static electiricity, malfunctioning electiral equipments, an open flame in contact with the liner or purposeful action. Interviews with engineers who were analyzing the risks led to estimates that such an explosion would occur with probability 0.25 as a result of static electricity, 0.20 as a result of malfunctioning electric equipment, 0.40 as a result of an open flame, and 0.75 result of purposeful action. These interviews also yielded subjective estimates of the probabilities of the four causes of 0.30, 0.40, 0.15 abd 0.15, respectively. Based on all information, what is the most likely cause of the explosion and what is the probability of explosion?
Let event A consist in the fact that an explosion will occur. Let's designate the causes:
"B_{1}" - static electiricity;
"B_{2}" - malfunctioning electiral equipments;
"B_{3}" - an open flame in contact with the liner;
"B_{4}" - purposeful action;
Then
"p({B_1}) = 0.3,\\,\\,p({B_2}) = 0.4,\\,p({B_3}) = p({B_4}) = 0.15" ,
"p(A|{B_1}) = 0.25,\\,\\,p({B_2}) = 0.2,\\,p(A|{B_3}) = 0.4,\\,p(A|{B_4}) = 0.75"
Then, according to the formula of total probability, the probability of explosion is
"p(A) = \\sum {p\\left( {{B_i}} \\right)} p(A|{B_i}) = 0.3 \\cdot 0.25 + 0.4 \\cdot 0.2 + 0.15 \\cdot 0.4 + 0.15 \\cdot 0.75 = 0.075 + 0.08 + 0.06 + {\\rm{0}}{\\rm{.1125 = 0}}{\\rm{.3275}}"
Using Bayes' formula, we find the probability of each cause:
"p({B_1}|A) = \\frac{{p({B_1})p(A|{B_1})}}{{p(A)}} = \\frac{{0.075}}{{0.3275}} \\approx 0.23"
"p({B_2}|A) = \\frac{{p({B_2})p(A|{B_2})}}{{p(A)}} = \\frac{{0.08}}{{0.3275}} \\approx 0.24"
"p({B_3}|A) = \\frac{{p({B_3})p(A|{B_3})}}{{p(A)}} = \\frac{{0.06}}{{0.3275}} \\approx 0.18"
"p({B_4}|A) = \\frac{{p({B_4})p(A|{B_4})}}{{p(A)}} = \\frac{{0.1125}}{{0.3275}} \\approx 0.34"
Then the most likely cause is purposeful action.
Answer: the probability of explosion is 0.3275, the most likely cause is purposeful action.
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