Advertisements
Advertisements
प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
Y | −1 | 0 | 1 |
P(Y) | 0.6 | 0.1 | 0.2 |
उत्तर १
P.m.f. of random variable should satisfy the following conditions :
(a) 0 ≤ pi ≤ 1
(b) ∑pi = 1
Y | −1 | 0 | 1 |
P(Y) | 0.6 | 0.1 | 0.2 |
Here ∑pi = 0.6 + 0.1 + 0.2
= 0.9 ≠ 1
Hence, P(Y) cannot be regarded as p.m.f. of the random variable Y.
उत्तर २
Here, pi > 0, `AA` i = 1, 2, 3
Now consider,
`sum_("i" = 1)^3 "P"_"i"` = 0.6 + 0.1 + 0.2
= 0.9 ≠ 1
∴ Given distribution is not p.m.f.
संबंधित प्रश्न
State if the following is not the probability mass function of a random variable. Give reasons for your answer.
X | 0 | 1 | 2 | 3 | 4 |
P(X) | 0.1 | 0.5 | 0.2 | − 0.1 | 0.2 |
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 |
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Determine:
- k
- P(X < 3)
- P( X > 4)
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the standard deviation of X.
The following is the p.d.f. of r.v. X:
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise.
Find P (x < 1·5)
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)`.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = `1/5`, for 0 ≤ x ≤ 5 and = 0 otherwise.
Find the probability that the waiting time is more than 4 minutes.
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:
Find expected value of and variance of X for the following p.m.f.
X | -2 | -1 | 0 | 1 | 2 |
P(x) | 0.3 | 0.3 | 0.1 | 0.05 | 0.25 |
Solve the following problem :
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p. m. f. of X.
The probability distribution of discrete r.v. X is as follows :
x = x | 1 | 2 | 3 | 4 | 5 | 6 |
P[x=x] | k | 2k | 3k | 4k | 5k | 6k |
(i) Determine the value of k.
(ii) Find P(X≤4), P(2<X< 4), P(X≥3).
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(0.5 ≤ x ≤ 1.5)
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)
F(x) is c.d.f. of discrete r.v. X whose distribution is
Xi | – 2 | – 1 | 0 | 1 | 2 |
Pi | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
Then F(– 3) = _______ .
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 r.v. X assumes values 1, 2, 3, ……. n with equal probabilities then E(X) = `("n" + 1)/(2)`
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 |
Determine the value of k.
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 :
Find the expected value and variance of the r. v. X if its probability distribution is as follows.
x | – 1 | 0 | 1 |
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 | 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 the p. m. f. of the r. v. X be
`"P"(x) = {((3 - x)/(10)", ","for" x = -1", "0", "1", "2.),(0,"otherwise".):}`
Calculate E(X) and Var(X).
If X denotes the number on the uppermost face of cubic die when it is tossed, then E(X) is ______
If a d.r.v. X takes values 0, 1, 2, 3, … with probability P(X = x) = k(x + 1) × 5–x, where k is a constant, then P(X = 0) = ______
If a d.r.v. X has the following probability distribution:
X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X = x) | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
then k = ______
The probability distribution of X is as follows:
X | 0 | 1 | 2 | 3 | 4 |
P(X = x) | 0.1 | k | 2k | 2k | k |
Find k and P[X < 2]
If p.m.f. of r.v. X is given below.
x | 0 | 1 | 2 |
P(x) | q2 | 2pq | p2 |
then Var(x) = ______
E(x) is considered to be ______ of the probability distribution of x.
The probability distribution of a discrete r.v.X is as follows.
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
P(X ≤ 4) = `square + square + square + square = square`
The probability distribution of a discrete r.v. X is as follows:
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | k | 2k | 3k | 4k | 5k | 6k |
- Determine the value of k.
- Find P(X ≤ 4)
- P(2 < X < 4)
- P(X ≥ 3)
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) = _______.
The probability distribution of X is as follows:
x | 0 | 1 | 2 | 3 | 4 |
P[X = x] | 0.1 | k | 2k | 2k | k |
Find
- k
- P[X < 2]
- P[X ≥ 3]
- P[1 ≤ X < 4]
- P(2)
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.
Given below is the probability distribution of a discrete random variable x.
X | 1 | 2 | 3 | 4 | 5 | 6 |
P(X = x) | K | 0 | 2K | 5K | K | 3K |
Find K and hence find P(2 ≤ x ≤ 3)