Advertisements
Advertisements
प्रश्न
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)
उत्तर
a. The table gives a probability distribution and therefore P[X = 0] + P[X = 1] + P[X = 2] + P[X = 3] + P[X = 4] = 1
i.e., 0.1 + k + 2k + 2k + k = 1
i.e., 6k = 0.9
∴ k = 0.15
k = 0.15
b. P[X < 2] = P[X = 0] + P[X = 1] = 0.1 + k
= 0.1 + 0.15
= 0.25
c. P[X ≥ 3] = P[X = 3] + P[X = 4] = 2k + k
= 3k
= 3(0.15)
= 0.45
d. P[1 ≤ X < 4] = P[X = 1] + P[X = 2] + P[X = 3]
= k + 2k + 2k
= 5k
= 5(0.15)
= 0.75
e. P(2) = P[X ≤ 2] = P[X = 0] + P[X = 1] + P[X = 2]
= 0.1 + k + 2k
= 0.1 + 3k
= 0.1 + 0.45
= 0.55
APPEARS IN
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 |
| P(X) | 0.4 | 0.4 | 0.2 |
Find expected value and variance of X for the following p.m.f.
| x | -2 | -1 | 0 | 1 | 2 |
| P(X) | 0.2 | 0.3 | 0.1 | 0.15 | 0.25 |
Find the mean number of heads in three tosses of a fair coin.
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)
The following is the p.d.f. of r.v. X :
f(x) = `x/8`, for 0 < x < 4 and = 0 otherwise
P ( 1 < x < 2 )
It is known that error in measurement of reaction temperature (in 0° c) in a certain experiment is continuous r.v. given by
f (x) = `x^2/3` , for –1 < x < 2 and = 0 otherwise
Find probability that X is negative
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 waiting time is between 1 and 3.
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 ) =
Solve the following :
The following probability distribution of r.v. X
| X=x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| P(X=x) | 0.05 | 0.1 | 0.15 | 0.20 | 0.25 | 0.15 | 0.1 |
Find the probability that
X is positive
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)
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).
Find the probability distribution of number of number of tails in three tosses of a coin
Find the probability distribution of number of heads in four tosses of a coin
Find expected value and variance of X, the number on the uppermost face of a fair die.
Given that X ~ B(n, p), if n = 10 and p = 0.4, find E(X) and Var(X)
Given that X ~ B(n,p), if n = 25, E(X) = 10, find p and Var (X).
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
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) = `k^4C_x` for x = 0, 1, 2, 3, 4 and P(x) = 0 otherwise then F(–1) = _______
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.
State whether the following is True or False :
| x | – 2 | – 1 | 0 | 1 | 2 |
| P(X = x) | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
If F(x) is c.d.f. of discrete r.v. X then F(–3) = 0
State whether the following is True or False :
If p.m.f. of discrete r.v. X is
| x | 0 | 1 | 2 |
| P(X = x) | q2 | 2pq | p2 |
then E(x) = 2p.
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 E(X) = 5 and Var(X) = 2.5, find n and p.
Find mean for the following probability distribution.
| X | 0 | 1 | 2 | 3 |
| P(X = x) | `1/6` | `1/3` | `1/3` | `1/6` |
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
k = `square`
Using the following activity, 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` |
Solution: µ = E(X) = `sum_("i" = 1)^3 x_"i""p"_"i"`
E(X) = `square + square + square = square`
Var(X) = `"E"("X"^2) - {"E"("X")}^2`
= `sum"X"_"i"^2"P"_"i" - [sum"X"_"i""P"_"i"]^2`
= `square - square`
= `square`
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.
