Question

Two numbers are drawn simultaneously from the set of integers from 1 to 12. If it is known that the sum of drawn two numbers is odd, then the probability that only one of the two numbers is a prime number, is

• A. \( \frac{5}{6} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{36}{53} \)
• D. \( \frac{11}{18} \)

Hint:

Sum of two integers is odd \(\iff\) one is even and one is odd.
List prime numbers carefully (1 is not prime).
For conditional probability \(P(B|A) = n(A \cap B)/n(A)\).

Answer 1

The Correct Option is D

Two functions \(f(x)\) and \(g(x)\) are linearly dependent if there exist constants \(c_1, c_2\), not both zero, such that \(c_1f(x) + c_2g(x) = 0\) for all \(x\). This is equivalent to one function being a constant multiple of the other (if neither is identically zero).

Let's examine option (b): \(f(x) = \sin x(4\sin^2 x - 3)\) and \(g(x) = \sin 3x\). 

We know the trigonometric identity for \(\sin 3x\): \(\sin 3x = 3\sin x - 4\sin^3 x\).

Now consider \(f(x)\): \(f(x) = \sin x(4\sin^2 x - 3) = 4\sin^3 x - 3\sin x\).

Comparing \(f(x)\) with \(\sin 3x\): \(f(x) = 4\sin^3 x - 3\sin x = -(3\sin x - 4\sin^3 x) = -\sin 3x\).

So, \(f(x) = -g(x)\), or \(f(x) + g(x) = 0\). Since we can write \(1 \cdot f(x) + 1 \cdot g(x) = 0\), with non-zero constants, the functions are linearly dependent.

Let's briefly check other options:

(a) \(e^x \sin 2x, e^x \cos 2x\): Linearly independent as \(\sin 2x\) and \(\cos 2x\) are independent.

(c) \(\cos x, x \cos x\): If \(c_1 \cos x + c_2 x \cos x = 0 \Rightarrow \cos x (c_1 + c_2 x) = 0\). For this to hold for all \(x\), \(c_1=0\) and \(c_2=0\). Linearly independent.

(d) \(e^{3x}, (x+1)e^{2x}\): Different exponential growth rates and forms. Linearly independent.

\[ \boxed{\sin x(4\sin^2 x - 3), \sin 3x} \]