Q2Advanced Engineering Mathematics

Question

A random variable x has the following probability distribution:

i) Find K. ii) Evaluate P (x < 6), P (x \geq 6) and P (0 < x < 5) iii) Find distribution function of x. iv) Find P (1.5 < x < 4.5 ∣ x > 2)

Answer 1

In probability theory, a discrete probability distribution characterizes the exact probabilities of occurrence for a countable number of distinct outcomes. The problem presents a random variable $X$ that can take integer values from 0 to 7, with their respective probabilities expressed in terms of an unknown constant $k$ . We must rigorously determine $k$ , analyze various probability subsets, and construct the cumulative distribution function (CDF).

Part I: Determining the Constant k

The most fundamental axiom of any valid probability mass function (PMF) is the Law of Total Probability, which states that the sum of the probabilities of all mutually exclusive and exhaustive events in the sample space must be exactly equal to 1. Mathematically:

x = 0 \sum 7 P (X = x) = 1

Substituting the given probability expressions for each value of $X$ :

P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) + P (7) = 1

0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + (7 k^{2} + k) = 1

Now, we group the like terms together. Grouping the linear $k$ terms: $k + 2 k + 2 k + 3 k + k = 9 k$ . Grouping the quadratic $k^{2}$ terms: $k^{2} + 2 k^{2} + 7 k^{2} = 10 k^{2}$ .

10 k^{2} + 9 k = 1 ⟹ 10 k^{2} + 9 k - 1 = 0

This is a standard quadratic equation. We can solve it by factoring the quadratic expression:

10 k^{2} + 10 k - k - 1 = 0

10 k (k + 1) - 1 (k + 1) = 0 ⟹ (10 k - 1) (k + 1) = 0

This yields two mathematically possible roots: $k = \frac{1}{10} = 0.1$ or $k = - 1$ . However, another fundamental axiom of probability theory dictates that all individual probabilities must be strictly non-negative ( $P (X = x) \geq 0$ for all $x$ ). If $k = - 1$ , then $P (1) = - 1$ , which is physically and mathematically impossible. Therefore, we rigidly reject the negative root.

Conclusion for Part I: $k = 0.1$

Part II: Evaluating Specific Probabilities

With $k = 0.1$ established, we can confidently calculate the specific numerical probability for any given event.

1. Evaluating $P (X < 6)$ : This represents the probability that the random variable $X$ takes any value strictly less than 6.

P (X < 6) = P (0) + P (1) + P (2) + P (3) + P (4) + P (5)

P (X < 6) = 0 + k + 2 k + 2 k + 3 k + k^{2} = 8 k + k^{2}

Substituting $k = 0.1$ : $8 (0.1) + (0.1)^{2} = 0.8 + 0.01 = 0.81$ .

2. Evaluating $P (X \geq 6)$ : This represents the probability that $X$ is 6 or greater. It can be calculated directly or by using the complement rule $1 - P (X < 6)$ . Let us calculate it directly for verification.

P (X \geq 6) = P (6) + P (7) = 2 k^{2} + (7 k^{2} + k) = 9 k^{2} + k

Substituting $k = 0.1$ : $9 (0.1)^{2} + 0.1 = 9 (0.01) + 0.1 = 0.09 + 0.1 = 0.19$ .

Verification: $P (X < 6) + P (X \geq 6) = 0.81 + 0.19 = 1.0$ , confirming our calculations are absolutely correct.

3. Evaluating $P (0 < X < 5)$ : This is the probability that $X$ is strictly between 0 and 5, exclusive.

P (0 < X < 5) = P (1) + P (2) + P (3) + P (4) = k + 2 k + 2 k + 3 k = 8 k

Substituting $k = 0.1$ : $8 (0.1) = 0.8$ .

Part III: Constructing the Cumulative Distribution Function (CDF)

The Cumulative Distribution Function, denoted as $F (x)$ , calculates the probability that the random variable $X$ will take a value less than or equal to $x$ . Mathematically, $F (x) = P (X \leq x)$ . Because $X$ is discrete, $F (x)$ is a step function.

$F (0) = P (0) = 0 = 0$
$F (1) = F (0) + P (1) = 0 + k = 0.1$
$F (2) = F (1) + P (2) = 0.1 + 2 k = 0.1 + 0.2 = 0.3$
$F (3) = F (2) + P (3) = 0.3 + 2 k = 0.3 + 0.2 = 0.5$
$F (4) = F (3) + P (4) = 0.5 + 3 k = 0.5 + 0.3 = 0.8$
$F (5) = F (4) + P (5) = 0.8 + k^{2} = 0.8 + 0.01 = 0.81$
$F (6) = F (5) + P (6) = 0.81 + 2 k^{2} = 0.81 + 0.02 = 0.83$
$F (7) = F (6) + P (7) = 0.83 + (7 k^{2} + k) = 0.83 + (0.07 + 0.1) = 1.0$

Part IV: Evaluating Conditional Probability

We are asked to evaluate $P (1.5 < X < 4.5 ∣ X > 2)$ . This is a conditional probability, which fundamentally measures the probability of an event occurring given that another event has already occurred. The standard formula is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ .

First, we must meticulously identify the discrete integer values of $X$ that satisfy the condition $A$ : $1.5 < X < 4.5$ . The valid integers are $X \in {2, 3, 4}$ .

Second, we identify the discrete integer values satisfying condition $B$ : $X > 2$ . The valid integers are $X \in {3, 4, 5, 6, 7}$ .

The intersection $A \cap B$ consists of the values that satisfy both conditions simultaneously. These are $X \in {3, 4}$ . Therefore, $P (A \cap B) = P (X = 3) + P (X = 4)$ .

The denominator is $P (X > 2) = 1 - P (X \leq 2)$ . Let us compute both:

P (A \cap B) = P (3) + P (4) = 2 k + 3 k = 5 k = 5 (0.1) = 0.5

P (B) = 1 - (P (0) + P (1) + P (2)) = 1 - (0 + 0.1 + 0.2) = 1 - 0.3 = 0.7

Finally, applying the conditional probability formula:

P (1.5 < X < 4.5 ∣ X > 2) = \frac{0.5}{0.7} = \frac{5}{7} \approx 0.7143

Q2Advanced Engineering Mathematics

Question

Answer

Part I: Determining the Constant k

Part II: Evaluating Specific Probabilities

Part III: Constructing the Cumulative Distribution Function (CDF)

Part IV: Evaluating Conditional Probability