Test: Probability- 1 - Computer Science Engineering (CSE) MCQ

Bag contains 10 white balls and 15 black balls. Required probability

P(atleast one of dice will have 6 facing 
= 1 - P(none of dice have 6 facing up)

The probability that the bottom card of a randomly shuffled deck is ace = 4/52 because there are 4 aces out of total 52 cards.
From the remaining 3 aces out of SVcards the probability that the top card is also an ace = 3/51.
So required probability 

Constraints are

Let the time taken for first and second modules be represented by x and y and total time = t.
∴ t = x + y is a random variable.
Now the joint density function

which is also called as convolution of f₁ and f₂, abbreviated as f₁ * f₂.

We know that:

Random variable assigns a real number to each possible outcome.

Let X be a discreet random variable,then

where V(x) is the variance of x,

Explanation:

• The difference between the expectation of the square of a random variable (E[X2]) and the square of the expectation of the random variable (E[X])2 is called the variance of a random variable
• Variance measure how far a set of numbers is spread out
• A variance of zero(R=0) indicates that all the values are identical

• A variance of X = R =E[X2]- (E[X])² This quantity is always non-negative as it is an expectation of a non-negative quantity
• A non-zero variance is always positive means R > 0

So, R ≥ 0 is the answer. Since variance is  and hence never negative, 

The p.d.t of the random variable is

The cumulative distribution function F(x) is the probability uptox as given below:

(a) is false since of Pand Q are independent

which need not be zero.
(b) is false since

(c) is false since independence and mutually exclusion are unrelated properties.
(d) is true