Given:

\( \vec{a} = \hat{i} - \hat{k}, \quad \) \(\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k},\) \( \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \)

Form the matrix:

\( M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} \)

Find the determinant:

\( \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 \)

Since the determinant is constant and non-zero, the vectors are linearly independent.

\( \boxed{\text{The matrix does not depend on } x \text{ or } y} \)

Quick Solution

Given: \( \vec{a}, \vec{b} \) are unit vectors and

\( 2\vec{a} + \vec{b} = 3 \)

Take magnitude on both sides:

\( |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 \)

Use identity:

\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

Set equal to 9:

\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]

Quick Solution

Given:

\( \int f(x)\, dx = g(x) \)

Required: \( \int x^5 f(x^3)\, dx \)

Use substitution:

Let \( u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} \)

Now rewrite the integral:

\[ \int x^5 f(x^3)\, dx = \int x^5 f(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int x^3 f(u)\, du \]

But \( x^3 = u \), so:

\[ \frac{1}{3} \int u f(u)\, du \]

Now integrate by parts or use the identity:

\[ \int u f(u)\, du = u g(u) - \int g(u)\, du \]

Final answer:

\[ \int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right] = x^3 g(x^3) - \int x^2 g(x^3)\, dx \]

\[ \boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx } \]

Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

If A ⊆ B and B ⊆ C, then find the cardinality of A ∪ B ∪ C.

Explanation

• Since A is a subset of B, everything in A is already in B.
• Since B is a subset of C, everything in B (and hence A) is already in C.
• Therefore, A ∪ B ∪ C = C.

✅ Final Answer

|A ∪ B ∪ C| = |C|

Probability — No Black Ball is Chosen

Given:

• Yellow balls = 5
• Black balls = 4
• Green balls = 3
• Total balls = 5 + 4 + 3 = 12

We are to find:

Probability that no black ball is selected when 3 balls are drawn at random.

Step 1: Total number of ways to choose any 3 balls from 12:

\[ \text{Total ways} = \binom{12}{3} = 220 \]

Step 2: Ways to choose 3 balls such that no black ball is chosen:

Only yellow and green balls are allowed ⇒ Total = 5 (yellow) + 3 (green) = 8
\[ \text{Favorable ways} = \binom{8}{3} = 56 \]

Step 3: Probability

\[ P(\text{no black ball}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{56}{220} = \frac{14}{55} \]

\[ \boxed{\text{Probability} = \frac{14}{55}} \]

Number of Roots

Given:

\( e^x \sin x = 1 \) has two real roots → say \( x_1 \) and \( x_2 \)

Apply Rolle’s Theorem:

Since \( f(x) = e^x \sin x \) is continuous and differentiable, and \( f(x_1) = f(x_2) \), ⇒ There exists \( c \in (x_1, x_2) \) such that \( f'(c) = 0 \)

Compute:

\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]

Correct Shortcut Method — Find \( f(5) \)

Step 1: Define a helper polynomial:

\[ g(x) = f(x) - (x + 1) \]

Given: \( f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0 \)

So, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]

Step 2: Use \( f(0) = 25 \) to find A:

\[ f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1 \]

Step 3: Compute \( f(5) \):

\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]

✅ Final Answer:   \( \boxed{f(5) = 30} \)

Maximum Value of \( f(x) = (x - 1)^2(x + 1)^3 \)

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let \( u = (x - 1)^2 \), \( v = (x + 1)^3 \)
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set \( f'(x) = 0 \): \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate \( f(x) \) at these points

• \( f(1) = 0 \)
• \( f(-1) = 0 \)
• \( f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 \)

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is \( \frac{3456}{3125} = 2^p \cdot 3^q / 3125 \)

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer:   \( \boxed{(p, q) = (7,\ 3)} \)

Sum of Complex Roots of Unity

Given:

\[ x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n} \]

Required: Find: \[ \sum_{k=1}^{n} x_k \]

This is the sum of all \( n^\text{th} \) roots of unity (from \( k = 1 \) to \( n \)).

We know: \[ \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 \] So shifting index from \( k = 1 \) to \( n \) just cycles the same roots: \[ \sum_{k=1}^{n} e^{2\pi i k/n} = 0 \]

✅ Final Answer:   \( \boxed{0} \)

Locus of Point on Parabola

Given: Point on parabola \( y^2 = 4a x \) is at distance \( 5a \) from focus \( (a, 0) \).

Distance Equation:

\[ (x - a)^2 + y^2 = 25a^2 \] \[ \Rightarrow (x - a)^2 + 4a x = 25a^2 \] \[ \Rightarrow x^2 + 2a x - 24a^2 = 0 \]

Solving gives: \( x = 4a \), \( y = 4a \)

✅ Final Answer: \( \boxed{(4a,\ 4a)} \)

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: \( \boxed{\frac{5}{18}} \)

Number of real solutions

Equation: $$\sin\!\left(e^x\right) \;=\; 5^x + 5^{-x}$$

Reasoning

• For all real \(x\), \(\sin(e^x)\in[-1,1]\).
• By AM–GM, \(5^x + 5^{-x} \ge 2\) (with equality only when \(5^x=5^{-x}\Rightarrow x=0\)).
• At \(x=0\): LHS \(=\sin(1)\approx 0.84\), RHS \(=2\) ⇒ not equal.
• Hence RHS is always \(\ge 2\) while LHS is always \(\le 1\) → they can never match.

✅ Conclusion

No real solution (number of real solutions = 0).

Multiply numerator and denominator by \(10^x\): \( y = \dfrac{10^{2x} - 1}{10^{2x} + 1} \).

Put \( t = 10^{2x} \), then \( y = \dfrac{t-1}{t+1} \). Solving, \( t = \dfrac{1+y}{1-y} \).

Hence, \( 10^{2x} = \dfrac{1+y}{1-y} \). Taking log base 10: \( 2x = \log_{10}\!\Big(\dfrac{1+y}{1-y}\Big) \).

✅ Final Answer

The inverse function is:
\[ f^{-1}(y) = \tfrac{1}{2}\,\log_{10}\!\left(\dfrac{1+y}{1-y}\right), \quad |y|<1 \]

Given: Foci are (4, 3) and (12, 5), and the ellipse passes through the origin (0, 0).

Step 1: Use ellipse definition

$PF_1 = \sqrt{(0 - 4)^2 + (0 - 3)^2} 

= \sqrt{25} 

= 5$

$PF_2 = \sqrt{(0 - 12)^2 + (0 - 5)^2} 

= \sqrt{169} 

= 13$

Total distance = $5 + 13 = 18 \Rightarrow 2a = 18 \Rightarrow a = 9$

Step 2: Distance between the foci

$2c = \sqrt{(12 - 4)^2 + (5 - 3)^2} = \sqrt{64 + 4} = \sqrt{68} \Rightarrow c = \sqrt{17}$

Step 3: Find eccentricity

$e = \dfrac{c}{a} = \dfrac{\sqrt{17}}{9}$

✅ Final Answer: $\boxed{\dfrac{\sqrt{17}}{9}}$

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

• $\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k} $

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

• Case 1: $-m + 5 = 4 \Rightarrow m = 1$
• Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Given: Vectors:

• $\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
• $\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
• $\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix} $

Step 2: Expand the determinant:

$ = \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2) $

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$, then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$

Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of \( a = 2^{1/6} \)

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity \( a^6 = 2 \Rightarrow a^5 = \frac{2}{a} \)

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$

Step 1: Recognize the series

The exponent is an infinite geometric series: $$ 1 + |\sin x| + |\sin x|^2 + |\sin x|^3 + \cdots $$

This is a geometric series with first term \( a = 1 \), common ratio \( r = |\sin x| \in [0,1] \), so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for \( x \in (-\pi, \pi) \)

We want all \( x \in (-\pi, \pi) \) such that \( |\sin x| = \frac{1}{2} \)

So \( \sin x = \pm \frac{1}{2} \). Within \( (-\pi, \pi) \), the values of \( x \) satisfying this are:

• $x = \frac{\pi}{6}$
• $x = \frac{5\pi}{6}$
• $x = -\frac{\pi}{6}$
• $x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Given System of Equations:

• $x + 2y + 2z = 5$
• $x + 2y + 3z = 6$
• $x + 2y + \lambda z = \mu$

Goal: Find values of $\lambda$ and $\mu$ such that the system has infinitely many solutions

Step 1: Write Augmented Matrix

$ [A|B] = \begin{bmatrix} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \end{bmatrix} $

Step 2: Row operations: Subtract $R_1$ from $R_2$ and $R_3$

$ \Rightarrow \begin{bmatrix} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \end{bmatrix} $

Step 3: For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < number of variables (3)

This happens when the third row becomes all zeros:

$ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 $

$\Rightarrow \lambda = 2,\quad \mu = 5$

✅ Final Answer: $\boxed{\lambda = 2,\ \mu = 5}$

Formula for work done:

\[ W = |F| \cdot |D| \cdot \cos\theta \]

Given:

• \( |F| = 40 \, \text{N} \)
• \( |D| = 3 \, \text{m} \)
• \( \theta = 60^\circ \)

Step 1: Plug in the values:

\[ W = 40 \cdot 3 \cdot \cos(60^\circ) \]

Step 2: Use \( \cos(60^\circ) = \frac{1}{2} \)

\[ W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J} \]

✅ Final Answer: $\boxed{60 \, \text{J}}$

Total people: 9

Married couple: 2 specific people among them

Total ways to choose 5 people from 9:

\[ \text{Total} = \binom{9}{5} = 126 \]

✅ Case 1: Both are selected

We fix the married couple (2 people), then choose 3 more from remaining 7:

\[ \binom{7}{3} = 35 \]

✅ Case 2: Both are NOT selected

We remove both from the pool, then choose 5 from remaining 7:

\[ \binom{7}{5} = \binom{7}{2} = 21 \]

✅ Favorable outcomes:

\[ \text{Favorable} = 35 + 21 = 56 \]

✅ Probability:

\[ \text{Required Probability} = \frac{56}{126} = \frac{28}{63} = \frac{4}{9} \]

✅ Final Answer: $\boxed{\dfrac{4}{9}}$

Using the set formula: \[ |T \cup C| = |T| + |C| - |T \cap C| \] \[ 200 = 120 + 144 - x \quad \Rightarrow \quad x = 64 \]

Minimum possible intersection: \[ x \geq |T| + |C| - 200 = 64 \] Maximum possible intersection: \[ x \leq \min(120,144) = 120 \]

✅ Final Answer

The range of \(x\) is: 64 ≤ x ≤ 120 Hence, \(m = 64, \; n = 120\).

Given:

• Number of patients (n): 3
• Probability of recovery (p): 0.6
• Probability of failure (q): 0.4
• We are asked to find: Probability that exactly 2 out of 3 patients will recover.

Solution: We will use the Binomial Probability Formula to solve this.

Step 1: Binomial Probability Formula:

The formula is:
$$ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r} $$ Where:
\( n = 3 \), \( r = 2 \), \( p = 0.6 \), \( 1 - p = 0.4 \)

Step 2: Calculate the binomial coefficient \( \binom{3}{2} \):

Step 3: Substitute values into the formula:

✅ Final Answer: 0.432

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

Given:

\[ \sin x = \sin y \quad \text{and} \quad \cos x = \cos y \]

✳ Step 1: Use the identity for sine

\[ \sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]

✳ Step 2: Use the identity for cosine

\[ \cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi \]

? Combine both conditions

For both \( \sin x = \sin y \) and \( \cos x = \cos y \) to be true, the only consistent solution is:

\[ x = y + 2n\pi \Rightarrow x - y = 2n\pi \]

✅ Final Answer:

\[ \boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}} \]

A speaks the truth in 40% of the cases and B in 50% of the cases.

What is the probability that they contradict each other while narrating an incident?

? Let’s Define:

• \( P(A_T) = 0.4 \) → A tells the truth
• \( P(A_L) = 0.6 \) → A lies
• \( P(B_T) = 0.5 \) → B tells the truth
• \( P(B_L) = 0.5 \) → B lies

? Contradiction happens in two cases:

• A tells the truth, B lies → \( 0.4 \times 0.5 = 0.2 \)
• A lies, B tells the truth → \( 0.6 \times 0.5 = 0.3 \)

Total probability of contradiction: \[ P(\text{Contradiction}) = 0.2 + 0.3 = \boxed{0.5} \]

✅ Final Answer:

\[ \boxed{\frac{1}{2}} \]

Statistics Puzzle: Variance under Linear Transformation

Given:

• Y = 10.5 + 2X
• Var(Y) = 100

Formula: If Y = a + bX, then Var(Y) = b² × Var(X)

Apply:

• 100 = 2² × Var(X)
• 100 = 4 × Var(X)
• Var(X) = 100 / 4 = 25

✅ Final Answer: The variance of the marks given by the first judge is 25.

A man starts at the origin \( O \), walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point \( P \). Find the displacement vector \( \vec{OP} \).

? Solution:

• North-East (45°): \[ \vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right) \]
• North-West (135°): \[ \vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right) \]
• Total Displacement: \[ \vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right) \]

✅ Final Answer:

\[ \boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) } \]