A ship is fitted with three engines E1, E2 and E3. The...

A ship is fitted with three engines $E_{1}, E_{2}$ and $E_{3}$. 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 $X$ denote the event that the ship is operational and let $X_{1}, X_{2}$ and $X_{3}$ denote respectively the events thatthe engines $E_{1}, E_{2}$ and $E_{3}$ are functioning. Which of the following is(are) true ? [2012] (a) $P\left[X_{1}^{c} \mid X\right]=\frac{3}{16}$ (b) $P$ [Exactly two engines of the ship are functioning $\mid X]=\frac{7}{8}$ (c) $P\left[X \mid X_{2}\right]=\frac{5}{16}$ (d) $P\left[X \mid X_{1}\right]=\frac{7}{16}$

A ship is fitted with three engines $E_{1}, E_{2}$ and $E_{3}$. 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 $X$ denote the event that the ship is operational and let $X_{1}, X_{2}$ and $X_{3}$ denote respectively the events thatthe engines $E_{1}, E_{2}$ and $E_{3}$ are functioning. Which of the following is(are) true ? [2012]
(a) $P\left[X_{1}^{c} \mid X\right]=\frac{3}{16}$
(b) $P$ [Exactly two engines of the ship are functioning $\mid X]=\frac{7}{8}$
(c) $P\left[X \mid X_{2}\right]=\frac{5}{16}$
(d) $P\left[X \mid X_{1}\right]=\frac{7}{16}$

Key Concepts

Conditional Probability

Conditional probability is the likelihood of an event happening given that another event has occurred. It is defined as P(A | B) = P(A ? B) / P(B) and is fundamental when the information about one part of an event influences the probability assessment of another, making it crucial for evaluating outcomes in interconnected scenarios.

Independent Events

Independent events are those whose occurrence does not affect the probability of one another. This property allows the probability of a combined event to be computed as the product of the probabilities of each individual event, simplifying calculations in multi-event systems.

Total Probability Theorem

The Total Probability Theorem provides a way to determine the probability of an event by partitioning the sample space into mutually exclusive cases and summing the probabilities of the event occurring in each case weighted by the probability of those cases. This theorem is valuable when dealing with complex systems where an event can occur in multiple distinct ways.

Event Combination Analysis

Event Combination Analysis involves evaluating the probabilities associated with various configurations of events, such as the chance of exactly k events occurring from a collection of independent events. This analysis is important when outcomes depend on the number of successes among several independent indicators, and it often utilizes combinatorial methods to systematically account for all possible configurations.

Transcript

00:01 Hello students our question is a variable x has probability distribution is x is equal to 4 8 and 12 and probability of x is equal to 1 out of 3 3 out of 5 1 out of 18 values of e of x and e of x squared respectively so first we will write our question so the question is our question is a random a variable x has probability distribution that is x is equal to 4 x is equal to 8 and x is equal to 2x the probability of x is equal to x is equal to 4 has probability of 1 third 8 has a probability of 3 5 and 12 has the probability of 1 out of 15 1 out of 15 so here here we have to find e of x and e of x y so let's start our solution this is our question the probability mass function is given to us so first we will find e of x is equal to as we know that e of x is equal to x into probability of x so here x is 4 and its probability is 1 out of 3 plus x is 8 and its probability is 3 out of 5 plus x is 12 and its probability is 1 out of so this will be equal to 4 out of 3 plus 24 out of 5 plus 12 plus 12 out of 15 so this will be equal to if we take 15 celsius, we get the 5 times 4 times 20 plus this is 5 not 15, sorry.

02:30 This is 5 not 15.

02:34 So 5, 3 times 24 times 324 times 72 plus 15, 1 10 15, 112 times 12.

02:47 This is equal to 1 04, 104 out of 15, and this will be our e of x.

03:00 Now, for e of x square, e of x square, as we know that this will be equal to x square into probability of x.

03:13 So first we will find the square of the x that is 4 square, 8 square and 12 square.

03:22 This will be equal to 4 square multiplied by probability is 1 out of 3 plus 8 square multiplied is 3 out of 5 plus 12 square multiplied by probability is 1 out of 5...

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