A ship is fitted with three engines E₁, E₂ and E₃. The engines function independently of each other

Q: Why do we need to calculate $P(X)$ first? Can't we just jump to the options?

Great question! In conditional probability, the denominator is always the probability of the condition (the event that has already happened). Since every option involves $P(X)$ or $P(X_i)$ in the denominator, knowing the total probability of the ship being operational ($P(X)$) is the foundation for everything else. It is like finding the total area of the sample space before you start calculating specific regions.

Q: Wait, why is $P(X_1^c \cap X)$ just $P(X_1^c \cap X_2 \cap X_3)$?

Think about the physical reality. $X$ means 'at least two engines work'. If $X_1$ has failed ($X_1^c$), the only way to have 'at least two' engines working is if both $X_2$ and $X_3$ are functioning. If only one of them worked, you would only have one engine total, which isn't enough to make the ship operational. So, $X_1^c \cap X$ is equivalent to $X_1^c \cap X_2 \cap X_3$.

Q: Why can't we just use energy conservation or some other shortcut?

Probability isn't about energy; it is about counting outcomes in a sample space. There are no shortcuts here because the events are independent but not equally likely. The most robust way to solve this is to list the mutually exclusive cases and sum their probabilities. It is tedious, but it is the only way to be 100% sure you haven't missed a scenario.

Question

A ship is fitted with three engines $E_1, E_2$ and $E_3$. The engines function independently of each other with respective probabilities $1/2, 1/4$ and $1/4$. For the ship to be operational at least two of its engines must function. Let $X$ denote the event that the ship is operational and let $X_1, X_2$ and $X_3$ denote respectively the events that the engines $E_1, E_2$ and $E_3$ are functioning. Which of the following is(are) true?

Options:

(A) $P[X_1^c | X] = 3/16$

(B) $P[	ext{Exactly two engines of the ship are functioning} | X] = 7/8$

(C) $P[X | X_2] = 5/16$

(D) $P[X | X_1] = 7/16$

Sigma Percentile · Accepted Answer

### The Ship and the Logic of Survival

Imagine you are the captain of a massive vessel navigating the open ocean. Your survival depends on three engines, $E_1$, $E_2$, and $E_3$.

These engines are independent, meaning the failure of one does not doom the others. We are given their probabilities of functioning: $P(X_1) = \frac{1}{2}$, $P(X_2) = \frac{1}{4}$, and $P(X_3) = \frac{1}{4}$.

Before we do anything else, we must acknowledge the failure rates—the complements. As a rule of thumb in JEE probability, always write down the failure probabilities immediately:

$$P(X_1^c) = 1 - \frac{1}{2} = \frac{1}{2}$$

$$P(X_2^c) = 1 - \frac{1}{4} = \frac{3}{4}$$

$$P(X_3^c) = 1 - \frac{1}{4} = \frac{3}{4}$$

These are your safety nets.

### Defining the Operational State ($X$)

The problem defines the ship as operational (Event $X$) if at least two engines function. This is the core geometric reality of the problem.

'At least two' is a classic phrase that splits into two distinct scenarios: either exactly two engines work, or all three engines work. To find the total probability $P(X)$, we must sum the probabilities of these mutually exclusive cases:

$$P(X) = P(X_1 X_2 X_3^c) + P(X_1 X_2^c X_3) + P(X_1^c X_2 X_3) + P(X_1 X_2 X_3)$$

Since the engines are independent, we calculate each term by multiplying the individual probabilities. Substituting our values, we get:

$$P(X) = \left( \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{3}{4} ight) + \left( \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{1}{4} ight) + \left( \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} ight) + \left( \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} ight)$$

Calculating these gives us:

$$\frac{3}{32} + \frac{3}{32} + \frac{1}{32} + \frac{1}{32} = \frac{8}{32} = \frac{1}{4}$$

The ship has a 25% chance of being operational. This is our anchor value.

### The Detective Work: Evaluating the Options

Now, we treat each option as a hypothesis to be tested using the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

**Option (A):** We test $P(X_1^c | X)$. This asks: if the ship is operational, what is the probability that engine 1 failed?

The numerator $P(X_1^c \cap X)$ represents the ship running on engines 2 and 3 only. That probability is $\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{32}$.

Dividing by $P(X) = \frac{1}{4}$, we get $\frac{1/32}{1/4} = \frac{1}{8}$. The option claims $\frac{3}{16}$, so **Option (A) is false**.

**Option (B):** We test $P(	ext{Exactly 2} | X)$. The intersection of 'exactly two' and 'operational' is just 'exactly two'.

We sum the first three cases from our $P(X)$ calculation: $\frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{7}{32}$.

Dividing by $P(X) = \frac{1}{4}$ gives $\frac{7/32}{1/4} = \frac{7}{8}$. This matches the option perfectly! **Option (B) is true**.

**Option (C):** We test $P(X | X_2)$. This is the probability the ship works given engine 2 is definitely working.

The numerator $P(X \cap X_2)$ includes cases where engine 2 works and the ship is operational: $(X_1 X_2 X_3^c)$, $(X_1^c X_2 X_3)$, and $(X_1 X_2 X_3)$. Summing these: $\frac{3}{32} + \frac{1}{32} + \frac{1}{32} = \frac{5}{32}$.

Dividing by $P(X_2) = \frac{1}{4}$ gives $\frac{5/32}{1/4} = \frac{5}{8}$. The option claims $\frac{5}{16}$, so **Option (C) is false**.

**Option (D):** Finally, $P(X | X_1)$. Similar to Option (C), we look for cases where engine 1 works and the ship is operational: $(X_1 X_2 X_3^c)$, $(X_1 X_2^c X_3)$, and $(X_1 X_2 X_3)$.

Summing these: $\frac{3}{32} + \frac{3}{32} + \frac{1}{32} = \frac{7}{32}$.

Dividing by $P(X_1) = \frac{1}{2}$ gives $\frac{7/32}{1/2} = \frac{7}{16}$. This matches the option! **Option (D) is true**.

By systematically breaking down the events, we have navigated the storm and found our way to the correct answers: **(B) and (D)**. Remember, in probability, clarity of cases is your best compass.

Select Answer:

Visualized Solution

The Ship and Its Engines

Probabilities of Failure

Condition for Operation (Event $X$ )

Calculating $P (X)$ - The Setup

Calculating $P (X)$ - Execution

Testing Option (A): $P (X_{1}^{c} ∣ X)$

Evaluating Option (A)

Testing Option (B): Exactly Two Engines

Evaluating Option (B)

Testing Option (C): $P (X ∣ X_{2})$

Evaluating Option (C)

Testing Option (D): $P (X ∣ X_{1})$

Evaluating Option (D)

Final Conclusion

The Sigma Insight: Conditional Probability

The Ship and the Logic of Survival

Defining the Operational State ( $X$ )

The Detective Work

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Let $E^{c}$ denote the complement of an event $E$ . Let $E, F, G$ be pairwise independent events with $P (G) > 0$ and $P (E \cap F \cap G) = 0$ . Then $P (E^{c} \cap F^{c} ∣ G)$ equals

Select Answer:

Visualized Solution

The Ship and Its Engines

Probabilities of Failure

Condition for Operation (Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X) - The Setup

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X) - Execution

Testing Option (A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X1c​∣X)