A ship is fitted with three engines E1, E2 and E3. The engines function independently of each other with respective probabilities \frac{1}{2} , . \frac{1}{4} \textrm{ }\textrm{ } and \textrm{ }\textrm{ } \frac{1}{4}. For the ship to be operational at least two of its engines must function. Let X denotes the event that the ship is operational and let X1, X2 and X3 denotes respectively the events that the engines E1, E2 and E3 are functioning. Which of the following is(are) true?

Engineering

Mathematics

Total Probability Theorem and Bayes Theorem

Question

A ship is fitted with three engines E₁, E₂ and E₃. The engines function independently of each other with respective probabilities $\frac{1}{2}, . \frac{1}{4} and \frac{1}{4}$ . For the ship to be operational at least two of its engines must function. Let X denotes the event that the ship is operational and let X₁, X₂ and X₃ denotes respectively the events that the engines E₁, E₂ and E₃ are functioning. Which of the following is(are) true?

$P [X | X_{1}] = \frac{7}{16}$

P [Exactly two engines of the ship are functioning | X] = $\frac{7}{8}$

${P[X|X}_{2}] = \frac{5}{16}$

$P [X_{1}^{c} | X] = \frac{3}{16}$

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P(X₁) = $\frac{1}{2}$ ; P(X₂) = $\frac{1}{4}$ and P(X₃) = $\frac{1}{4}$

P(X) = P (atleast 2 of X₁, X₂, X₃ happens)

= P(X₁ $\cap$ X₂) + P(X₂ $\cap$ X₃) + P(X₃ $\cap$ X₁) – 2P(X₁ $\cap$ X₂ $\cap$ X₃)

$= \frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{2} - 2 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{4}$

(A) $P (\frac{X_{1}^{c}}{X}) = \frac{P (X_{1}^{c} \cap X)}{P (X)} = \frac{P (X_{1}^{c} \cap X_{3} \cap X_{2})}{P (X)} = \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}} = \frac{1}{8}$

(B) $\frac{P (Exactly 2 of X_{1}, X_{2}, X_{3} happens)}{X} = \frac{P ((Exactly 2 of X_{1} X_{2} X_{3}) \cap X)}{P (X_{1})} = \frac{\frac{1}{4} - \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4}}{\frac{1}{4}} = \frac{7}{8}$

(C) $P (\frac{X}{X_{2}}) = \frac{P (X \cap X_{2})}{P (X_{2})} = \frac{P (X_{1} \cap X_{2} \cap X_{3}) + P (X_{2} \cap X_{1}^{c} \cap X_{3}) + P (X_{2} \cap X_{1} \cap X_{3}^{c})}{P (X_{2})}$

$= \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{2} \times \frac{3}{4}}{\frac{1}{4}} = \frac{5}{8}$

$P (\frac{X}{X_{1}}) = \frac{P (X \cap X_{1})}{P (X_{1})} = \frac{P (X_{1} \cap X_{2} \cap X_{3}) + P (X_{1} \cap X_{2}^{c} \cap X_{3}) + P (X_{1} \cap X_{2} \cap X_{3}^{c})}{P (X_{1})}$

$= \frac{\frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} + \frac{1}{2} \times \frac{3}{4} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{4} \times \frac{3}{4}}{\frac{1}{2}} = \frac{7}{16}$