Let U and V be two independent zero mean Gaussian random variables of variances 14 and 19 respectively- The probability P3V- 2U is

Let nbsp; nbsp; nbsp;x = 3V 2u then nbsp; nbsp; E(x) = E(3V 2u) = 3 × 0 2 × 0 = 0 So nbsp; nbsp; nbsp;Z_x=x μσ=0 0σ=0] Hence, P(3V ...

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Engineering Mathematics

Normal Distribution Properties

Let U and V b...

Question

Let U and V be two independent zero mean Gaussian random variables of variances $\frac{1}{4} and \frac{1}{9}$ respectively. The probability $P (3 V \geq 2 U) is$

\frac{4}{9}

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\frac{1}{2}

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\frac{2}{3}

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\frac{5}{9}

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Solution

The correct option is B $\frac{1}{2}$
Let x = 3V - 2u

then E(x) = E(3V - 2u) = 3 × 0 - 2 × 0 = 0

So $Z_{x} = \frac{x - μ}{σ} = \frac{0 - 0}{σ} = 0$ ]

$Hence, P (3 V \geq 2 u) = P (3 V - 2 u \geq 0) = P (x \geq 0)$

$= P (z_{x} \geq 0) = \frac{1}{2}$

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Let U and V be two independent zero mean Gaussian random variables of variances 14 and 19 respectively. The probability P(3V≥2U) is

The correct option is B 12Let x = 3V - 2u then E(x) = E(3V - 2u) = 3 × 0 - 2 × 0 = 0 So Zx=x−μσ=0−0σ=0] Hence, P(3V≥2u)=P(3V−2u≥0)=P(x≥0) =P(zx≥0)=12

Let U and V be two independent zero mean Gaussian random variables of variances $\frac{1}{4} and \frac{1}{9}$ respectively. The probability $P (3 V \geq 2 U) is$

The correct option is B $\frac{1}{2}$
Let x = 3V - 2u

then E(x) = E(3V - 2u) = 3 × 0 - 2 × 0 = 0

So $Z_{x} = \frac{x - μ}{σ} = \frac{0 - 0}{σ} = 0$ ]

$Hence, P (3 V \geq 2 u) = P (3 V - 2 u \geq 0) = P (x \geq 0)$

$= P (z_{x} \geq 0) = \frac{1}{2}$