हिंदी

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) = - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

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) =

विकल्प

  • `7/25`

  • `16/25`

  • `18/25`

  • `19/25`

MCQ
रिक्त स्थान भरें

उत्तर

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) = `16/25`

Hint : `k[1/5^0 + 2/5^1 + 3/5^2 + ....]=1`

`Let s = k/5^0+ 2k/5^1+3k/5^2+ ...............`

i.e `s = k + 2k/5+3k/5^2+..........`

∴`1/5s = k/5 + (2k)/5^2 + (3k)/5^3 + ........`

∴ `s - 1/5s = k + k/5 + k/5^2 + k/5^3 + ..........`

∴ `4/5s=k[1 + 1/5 + 1/5^2 + 1/5^3 + ....]`

= `k [1/(1 - 1/5)] = 5k/4`

∴ `s = (25k)/16 = 1`

∴ `k = 16/25`

∴`P(x=0) = k(0+1)5^0 = k = 16/25.`

shaalaa.com
Probability Distribution of Discrete Random Variables
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Probability Distributions - Miscellaneous Exercise 1 [पृष्ठ २४२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 7 Probability Distributions
Miscellaneous Exercise 1 | Q 4 | पृष्ठ २४२

संबंधित 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

State if the following is not the probability mass function of a random variable. Give reasons for your answer.

0 -1 -2
P(X) 0.3 0.4 0.3

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)


Find k if the following function represent p.d.f. of r.v. X

f (x) = kx, for 0 < x < 2 and = 0 otherwise, Also find P `(1/ 4 < x < 3 /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 p.m.f. of a d.r.v. X is P (X = x) = `x^2 /(n (n + 1))`, for x = 1, 2, 3, . . ., n and = 0, otherwise 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) =


Solve the following :

Identify the random variable as either discrete or continuous in each of the following. Write down the range of it.

Amount of syrup prescribed by physician.


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(x≤1)


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)


70% of the members favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and Var(X).


Find k if the following function represents the p. d. f. of a r. v. X.

f(x) = `{(kx,  "for"  0 < x < 2),(0,  "otherwise."):}`

Also find `"P"[1/4 < "X" < 1/2]`


Given that X ~ B(n,p), if n = 25, E(X) = 10, find p and Var (X).


X is r.v. with p.d.f. f(x) = `"k"/sqrt(x)`, 0 < x < 4 = 0 otherwise then x 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) = `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.m.f. of discrete r.v. X is

x 0 1 2
P(X = x) q2 2pq p2 

then E(x) = 2p.


If r.v. X assumes values 1, 2, 3, ……. n with equal probabilities then E(X) = `("n" + 1)/(2)`


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 :

Let X∼B(n,p) If n = 10 and E(X)= 5, find p and Var(X).


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 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) = ______


If the p.m.f. of a d.r.v. X is P(X = x) = `{{:(("c")/x^3",", "for"  x = 1","  2","  3","),(0",", "otherwise"):}` then E(X) = ______


If 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) is ______


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 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`


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

  1. k
  2. P[X < 2]
  3. P[X ≥ 3]
  4. P[1 ≤ X < 4]
  5. P(2)

The value of discrete r.v. is generally obtained by counting.


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)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×