Probability and Statistics MCQs Question and Answer in Engineering Maths

Let the probability density function of a random variable, X, be given as: $${{\text{f}}_{\text{x}}}\left( {\text{x}} \right) = \frac{3}{2}{{\text{e}}^{ - 3{\text{x}}}}{\text{u}}\left( {\text{x}} \right) + {\text{a}}{{\text{e}}^{4{\text{x}}}}{\text{u}}\left( { - {\text{x}}} \right)$$      where u(x) is the unit step function. Then the value of 'a' and Prob{X ≤ 0}, respectively, are

A random variable is uniformly distributed over the interval 2 to 10. Its variance will be

A box has 8 red balls and 8 green balls. Two balls are drawn randomly in succession from the box without replacement. The probability that the first ball drawn is red and the second ball drawn is green is

The value of the integral $$I = \frac{1}{{\sqrt {2\pi } }}\int\limits_0^\infty {\exp \left( { - \frac{{{{\text{x}}^2}}}{8}} \right){\text{dx}}} $$      is

The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurement more than 10.15 mm will be

For the function f(x) = a + bx, 0 ≤ x ≤ 1, to be a valid probability density function, which one of the following statements is correct?

A continuous random variable X has a probability density f(x) = e^-x, 0 < x < $$\infty $$ . Then P{X > 1} is

Two white and two black balls, kept in two bins are arranged in four ways as shown below. In each arrangement, a bin has to be chosen randomly and only one ball needs to be picked randomly from the chosen bin. Which one of the following arrangements has the highest probability for getting a white face picked?

If a random variable X satisfies the Poisson's distribution with a mean value of 2, then probability that X ≥ 2 is

If probability density function of a random variable x is
f(x) = x² for -1 ≤ x ≤ 1, and
       = 0 for any other value of x
then, the percentage probability $${\text{P}}\left( { - \frac{1}{3} \leqslant {\text{x}} \leqslant \frac{1}{3}} \right)$$    is