Advertisements
Advertisements
प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
X | 0 | -1 | -2 |
P(X) | 0.3 | 0.4 | 0.3 |
उत्तर
P.m.f. of random variable should satisfy the following conditions :
(a) 0 ≤ pi ≤ 1
(b) ∑pi = 1.
X | 0 | -1 | -2 |
P(X) | 0.3 | 0.4 | 0.3 |
(a) Here 0 ≤ pi ≤ 1
(b) ∑pi = 0.3 + 0.4 + 0.3 = 1
Hence, P(X) can be regarded as p.m.f. of the random variable X.
APPEARS IN
संबंधित प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer
Z | 3 | 2 | 1 | 0 | −1 |
P(Z) | 0.3 | 0.2 | 0.4 | 0 | 0.05 |
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
P(x > 2)
Find k, if the following function represents p.d.f. of r.v. X.
f(x) = kx(1 – x), for 0 < x < 1 and = 0, otherwise.
Also, find `P(1/4 < x < 1/2) and P(x < 1/2)`.
Choose the correct option from the given alternative:
If a d.r.v. X takes values 0, 1, 2, 3, . . . which probability P (X = x) = k (x + 1)·5 −x , where k is a constant, then P (X = 0) =
Choose the correct option from the given alternative :
If p.m.f. of a d.r.v. X is P (x) = `c/ x^3` , for x = 1, 2, 3 and = 0, otherwise (elsewhere) then E (X ) =
Choose the correct option from the given alternative:
If the a d.r.v. X has the following probability distribution :
x | -2 | -1 | 0 | 1 | 2 | 3 |
p(X=x) | 0.1 | k | 0.2 | 2k | 0.3 | k |
then P (X = −1) =
Choose the correct option from the given alternative:
If the a d.r.v. X has the following probability distribution :
x | -2 | -1 | 0 | 1 | 2 | 3 |
p(X=x) | 0.1 | k | 0.2 | 2k | 0.3 | k |
then P (X = −1) =
The following is the c.d.f. of r.v. X
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
F(X) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 |
*1 |
P (–1 ≤ X ≤ 2)
The following is the c.d.f. of r.v. X
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
F(X) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 |
1 |
P (X ≤ 3/ X > 0)
Let X be amount of time for which a book is taken out of library by randomly selected student and suppose X has p.d.f
f (x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise. Calculate: P(x ≥ 1.5)
Find the probability distribution of number of number of tails in three tosses of a coin
Find expected value and variance of X, the number on the uppermost face of a fair die.
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x E(X) = _______
Choose the correct alternative :
If X ∼ B`(20, 1/10)` then E(X) = _______
Fill in the blank :
If X is discrete random variable takes the value x1, x2, x3,…, xn then \[\sum\limits_{i=1}^{n}\text{P}(x_i)\] = _______
If F(x) is distribution function of discrete r.v.x with p.m.f. P(x) = `(x - 1)/(3)` for x = 0, 1 2, 3, and P(x) = 0 otherwise then F(4) = _______.
State whether the following is True or False :
If P(X = x) = `"k"[(4),(x)]` for x = 0, 1, 2, 3, 4 , then F(5) = `(1)/(4)` when F(x) is c.d.f.
Solve the following problem :
The probability distribution of a discrete r.v. X is as follows.
X | 1 | 2 | 3 | 4 | 5 | 6 |
(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Find P(X ≤ 4), P(2 < X < 4), P(X ≤ 3).
Solve the following problem :
The p.m.f. of a r.v.X is given by
`P(X = x) = {(((5),(x)) 1/2^5", ", x = 0", "1", "2", "3", "4", "5.),(0,"otherwise"):}`
Show that P(X ≤ 2) = P(X ≤ 3).
Solve the following problem :
The following is the c.d.f of a r.v.X.
x | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |
F (x) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 | 1 |
Find the probability distribution of X and P(–1 ≤ X ≤ 2).
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
x | 1 | 2 | 3 |
P(X = x) | `(1)/(5)` | `(2)/(5)` | `(2)/(5)` |
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
x | 1 | 2 | 3 | ... | n |
P(X = x) | `(1)/"n"` | `(1)/"n"` | `(1)/"n"` | ... | `(1)/"n"` |
Solve the following problem :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
X | 0 | 1 | 2 | 3 | 4 | 5 |
P(X = x) | `(1)/(32)` | `(5)/(32)` | `(10)/(32)` | `(10)/(32)` | `(5)/(32)` | `(1)/(32)` |
Solve the following problem :
Let X∼B(n,p) If n = 10 and E(X)= 5, find p and Var(X).
The p.m.f. of a d.r.v. X is P(X = x) = `{{:(((5),(x))/2^5",", "for" x = 0"," 1"," 2"," 3"," 4"," 5),(0",", "otherwise"):}` If a = P(X ≤ 2) and b = P(X ≥ 3), then
If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(x/("n"("n" + 1))",", "for" x = 1"," 2"," 3"," .... "," "n"),(0",", "otherwise"):}`, then E(X) = ______
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as number greater than 4 appears on at least one die.
The values of discrete r.v. are generally obtained by ______
If X is discrete random variable takes the values x1, x2, x3, … xn, then `sum_("i" = 1)^"n" "P"(x_"i")` = ______
The value of discrete r.v. is generally obtained by counting.
The p.m.f. of a random variable X is as follows:
P (X = 0) = 5k2, P(X = 1) = 1 – 4k, P(X = 2) = 1 – 2k and P(X = x) = 0 for any other value of X. Find k.