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Animated Solution for Mathematics - Probability: Let a random variable $X$ take values $0, 1, 2, 3$ with $P (X = 0) = P (X = 1) = p$ , $P (X = 2) = P (X = 3)$ and $E (X^{2}) = 2 E (X)$ . Then the value of $8 p - 1$ is :

Q: I keep getting confused between $E(X^2)$ and $(E(X))^2$. Are they the same?

They are definitely not the same! $E(X^2)$ is the 'second moment', calculated by squaring the values of $X$ first and then finding the weighted average. $(E(X))^2$ is the square of the mean. The difference between these two, $E(X^2) - (E(X))^2$, is actually the definition of the variance of $X$.

Q: Why did we introduce the variable $q$ instead of just using $P(X=2)$?

It is a classic problem-solving strategy! Introducing $q$ simplifies the algebra significantly. By replacing $P(X=2)$ and $P(X=3)$ with a single variable, we reduce the complexity of our equations, making it much easier to spot the relationship between $p$ and $q$.

Select Answer:

Visualized Solution

Introduction to the Random Variable

Random Variable $X \in {0, 1, 2, 3}$
We are given a probability distribution for $X$ .

Defining the Probabilities

Given: $P (X = 0) = P (X = 1) = p$
Given: $P (X = 2) = P (X = 3)$
Let $P (X = 2) = P (X = 3) = q$

The Fundamental Law of Probability

The sum of all probabilities in a valid distribution must equal $1$ .
$\sum P (X = x_{i}) = 1$

Establishing the First Equation

$P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1$
$p + p + q + q = 1$
$2 p + 2 q = 1 \Rightarrow p + q = \frac{1}{2}$

Defining Expected Value $E (X)$

Expected Value (Mean) formula:
$E (X) = \sum x_{i} P (X = x_{i})$

Calculating $E (X)$

$E (X) = 0 \cdot p + 1 \cdot p + 2 \cdot q + 3 \cdot q$
$E (X) = 0 + p + 2 q + 3 q$
$E (X) = p + 5 q$

Defining the Second Moment $E (X^{2})$

Second Moment formula:
$E (X^{2}) = \sum x_{i}^{2} P (X = x_{i})$

Calculating $E (X^{2})$

$E (X^{2}) = 0^{2} \cdot p + 1^{2} \cdot p + 2^{2} \cdot q + 3^{2} \cdot q$
$E (X^{2}) = 0 + 1 \cdot p + 4 \cdot q + 9 \cdot q$
$E (X^{2}) = p + 13 q$

Applying the Given Condition

Given Condition: $E (X^{2}) = 2 E (X)$
Substitute the derived expressions:
$p + 13 q = 2 (p + 5 q)$

Simplifying the Master Equation

Expand the right side:
$p + 13 q = 2 p + 10 q$
Rearrange terms:
$13 q - 10 q = 2 p - p$
$3 q = p \Rightarrow q = \frac{p}{3}$

Solving for $p$

Recall from earlier: $p + q = \frac{1}{2}$
Substitute $q = \frac{p}{3}$ :
$p + \frac{p}{3} = \frac{1}{2}$

Final Calculation for $p$

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

The Way Forward: Finding $8 p - 1$

We need to find the value of $8 p - 1$ .
Substitute $p = \frac{3}{8}$ :
$8 (\frac{3}{8}) - 1$
$= 3 - 1 = 2$
Final Answer: 2

00:00 / 00:00

The Sigma Insight: Random Variables and Probability Distributions

Solution Diagram

Analyzing the Setup

Imagine a system where a random variable

X

can take four distinct values:

0, 1, 2,

and

3

. To keep our thoughts organized, we create a probability distribution table.

We are given that

P (X = 0) = P (X = 1) = p

. For the remaining values, we define

P (X = 2) = P (X = 3) = q

The Law of Totality

In the world of probability, the sum of all probabilities in a valid distribution must equal

1

. Mathematically, we write this as

\sum P (X = x_{i}) = 1

Applying this to our specific case, we sum the probabilities for all possible values of

X

P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

Substituting our variables, we get

p + p + q + q = 1

, which simplifies to

2 p + 2 q = 1

. Dividing by

2

, we arrive at our first vital relationship:

p + q = \frac{1}{2}

The Moments of Truth

Now, we enter the realm of 'moments'. The expected value, or mean,

E (X)

, is the weighted average of the values of

X

defined by

E (X) = \sum x_{i} P (X = x_{i})

Calculating this, we find:

E (X) = 0 \cdot p + 1 \cdot p + 2 \cdot q + 3 \cdot q = p + 5 q

Next, we tackle the second moment,

E (X^{2})

. We must square the values of

X

before multiplying by their probabilities:

E (X^{2}) = 0^{2} \cdot p + 1^{2} \cdot p + 2^{2} \cdot q + 3^{2} \cdot q = p + 13 q

The Algebraic Bridge

The problem provides us with the condition

E (X^{2}) = 2 E (X)

. We substitute our expressions for

E (X)

and

E (X^{2})

into this condition:

p + 13 q = 2 (p + 5 q)

Expanding the right side, we get

p + 13 q = 2 p + 10 q

. By rearranging the terms, we find

3 q = p

, or

q = \frac{p}{3}

Final Calculation

We return to our first equation,

p + q = \frac{1}{2}

. Substituting

q = \frac{p}{3}

into this, we get:

p + \frac{p}{3} = \frac{1}{2} \Rightarrow \frac{4 p}{3} = \frac{1}{2}

Solving for

p

, we multiply both sides by

\frac{3}{4}

to get

p = \frac{3}{8}

. Finally, the question asks for the value of

8 p - 1

Substituting our value of

p

8 (\frac{3}{8}) - 1 = 3 - 1 = 2

The final answer is 2.

Common Traps & FAQs

Wait, why is the sum of probabilities equal to 1? Can't it be anything else?

That is a fundamental axiom of probability theory! In any discrete probability distribution, the random variable

X

must take one of its possible values with certainty. Therefore, the sum of the probabilities of all mutually exclusive outcomes must be exactly 1. If it were not 1, the distribution would be invalid.

I keep getting confused between

E (X^{2})

and

(E (X))^{2}

. Are they the same?

They are definitely not the same!

E (X^{2})

is the 'second moment', calculated by squaring the values of

X

first and then finding the weighted average.

(E (X))^{2}

is the square of the mean. The difference between these two,

E (X^{2}) - (E (X))^{2}

, is actually the definition of the variance of

X

Why did we introduce the variable

q

instead of just using

P (X = 2)

It is a classic problem-solving strategy! Introducing

q

simplifies the algebra significantly. By replacing

P (X = 2)

and

P (X = 3)

with a single variable, we reduce the complexity of our equations, making it much easier to spot the relationship between

p

and

q

Animated Solution for Mathematics - Probability: Let a random variable $X$ take values $0, 1, 2, 3$ with $P (X = 0) = P (X = 1) = p$ , $P (X = 2) = P (X = 3)$ and $E (X^{2}) = 2 E (X)$ . Then the value of $8 p - 1$ is :

Select Answer:

Visualized Solution

Introduction to the Random Variable

Defining the Probabilities

The Fundamental Law of Probability

Establishing the First Equation

Defining Expected Value $E (X)$

Calculating $E (X)$

Defining the Second Moment $E (X^{2})$

Calculating $E (X^{2})$

Applying the Given Condition

Simplifying the Master Equation

Solving for $p$

Final Calculation for $p$

The Way Forward: Finding $8 p - 1$

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Law of Totality

The Moments of Truth

The Algebraic Bridge

Final Calculation

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Select Answer:

Visualized Solution

Introduction to the Random Variable

Defining the Probabilities

The Fundamental Law of Probability

Establishing the First Equation

Defining Expected Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X)

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

Defining the Second Moment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X2)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E(X2)

Applying the Given Condition

Simplifying the Master Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

The Way Forward: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 8p−1

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Law of Totality

The Moments of Truth

The Algebraic Bridge

Final Calculation

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