$S = \{0, 1, 2, 3\}$, being the number of individuals of a species within a habitat.
Consider the probability distribution of X with $Pr(X = 0) = 0.15$,
$Pr(X = 1) = 0.25$ and $Pr(X = 3) = 0.5$, where Pr denotes probability. The value of
$Pr(X = 2)$ is __________. (Round off to two decimal places)
We are given a discrete random variable X representing the number of individuals of a species in a habitat. The possible values for X are $S = \{0, 1, 2, 3\}$.
We need to find the probability $Pr(X = 2)$.
A fundamental property of any probability distribution is that the sum of probabilities for all possible values of the random variable must equal 1.
Mathematically, this is expressed as:
$ \sum_{x \in S} Pr(X = x) = 1 $Substituting the possible values of X:
$ Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) = 1 $Now, let's plug in the known probability values:
$ 0.15 + 0.25 + Pr(X = 2) + 0.5 = 1 $Combine the known probabilities:
$ 0.90 + Pr(X = 2) = 1 $To find $Pr(X = 2)$, subtract 0.90 from both sides of the equation:
$ Pr(X = 2) = 1 - 0.90 $ $ Pr(X = 2) = 0.10 $The calculated value for $Pr(X = 2)$ is 0.10. This value falls within the specified range of 0.05 to 0.15.
Consider the following probability density function for a random variable x.
(a) \(f_{1}(x)=1\ ;\ -1\le x\le 1\)
(b) \(f_{2}(x)=x\ ;\ 0\le x\le 1\)
(c) \(f_{3}(x)=(1-x)\ ;\ -1\le x\le 1\)
Arrange the above functions in terms of the increasing value of mean of random variable x.
Considering all the symbols with their usual meanings, match the following :
| List - I | List - II |
| (a) \(f_{X}(x)\) | (i) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx\) |
| (b) \(\displaystyle\int_{-\infty}^{\infty}f_{X}(x)\,dx\) | (ii) \(\displaystyle\int_{-\infty}^{a}f_{X}(x)\,dx\) |
| (c) \(F_{X}(a)\) | (iii) \(\displaystyle\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dy\) |
| (d) \(f_{Y}(y)\) | (iv) 1 |
Codes :
The density function of a random variable is given by \(p(x)=Ke^{-\frac{x^{2}}{2}}\) for \(-\infty \lt x \lt \infty\). The value of K should be :
If the odds in favour of any random event A are 5 ∶ 6, then the odds against the event are:
If random variable X follows binomial distribution with parameter n and p with mean 15 and variance 10, then the value of mode is