Answer to Question #193908 in Statistics and Probability for Su su

Question

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?

Answer 1

Given that an explosion in a liquefied natural gas tank undergoing repair has occurred. Interviews with engineers who were analyzing the risks led to the following estimates. The probability that the explosion would occur as a result of static electricity is 0.25, as a result of open flame is 0.40, as a result of malfunctioning of electric equipment is 0.20, as a result of purposeful action is 0.75. The subjective estimates of four causes are 0.30, 0.40, 0.15 and 0.15 respectively. We have to find the most likely cause of the explosion.

We know that if $B_1, B_2,...B_k$ are mutually exclusive events of which one must occur then

$P(B_i|A) = \frac{P(B_i)P(A|B_i)}{P(B_1)P(A_|B_1) + P(B_2)P(A_|B_2)+...+P(B_k)P(A_|B_k)}$ for i = 1,2,...k

Let A denote the explosion occurred and $B_1, B_2, B_3, B_4$ denote the events that the explosion occurred as a result of static electricity, open flame, malfunctioning of electric equipment and purposeful action respectively.

Given

$P(B_1) = 0.25 \\ P(B_2) = 0.40 \\ P(B_3)=0.20 \\ P(B_4) = 0.75 \\ P(A|B_1) = 0.30 \\ P(A|B_2) = 0.15 \\ P(A|B_3) = 0.40 \\ P(A|B_4) = 0.15$

We find the probability that the explosion occurred as a result of each of the four causes. Probability that the explosion occurred as a result of static electricity is given by:

$P(B_1|A) = \frac{P(B_1)P(A|B_1)}{P(B_1)P(A_|B_1) + P(B_2)P(A_|B_2)+...+P(B_4)P(A_|B_4)} \\ = \frac{0.25 \times 0.30}{ 0.25 \times 0.30 + 0.40 \times 0.15 + 0.20 \times 0.40 + 0.75 \times 0.15 } \\ = 0.229$

Probability that the explosion occurred as a result of open flame is given by:

$P(B_2|A) = \frac{P(B_2)P(A|B_2)}{P(B_1)P(A_|B_1) + P(B_2)P(A_|B_2)+...+P(B_4)P(A_|B_4)} \\ = \frac{0.40 \times 0.15}{ 0.25 \times 0.30 + 0.40 \times 0.15 + 0.20 \times 0.40 + 0.75 \times 0.15 } \\ = 0.183$

Probability that the explosion occurred as a result of electric equipment is given by:

$P(B_3|A) = \frac{P(B_3)P(A|B_3)}{P(B_1)P(A_|B_1) + P(B_2)P(A_|B_2)+...+P(B_4)P(A_|B_4)} \\ = \frac{0.20 \times 0.40 }{0.25 \times 0.30 + 0.40 \times 0.15 + 0.20 \times 0.40 + 0.75 \times 0.15} \\ = 0.244$

Probability that the explosion occurred as a result of electric equipment is given by:

$P(B_4|A) = \frac{P(B_4)P(A|B_4)}{P(B_1)P(A_|B_1) + P(B_2)P(A_|B_2)+...+P(B_4)P(A_|B_4)} \\ = \frac{0.75+0.15}{0.25 \times 0.30 + 0.40 \times 0.15 + 0.20 \times 0.40 + 0.75 \times 0.15} \\ = 0.344$