English

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. - Mathematics and Statistics

Advertisements
Advertisements

Question

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.

Sum
Advertisements

Solution 1

When a die is tossed twice, the sample space S has 6 × 6 = 36 sample points.

∴ n(S) = 36

Trial will be a success if the number on at least one die is 5 or 6.

Let X denote the number of dice on which 5 or 6 appears.

Then X can take values 0, 1, 2

When X = 0 i.e., 5 or 6 do not appear on any of the dice, then

X = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}.

∴ n(X) = 16

∴ P(X = 0) =`(n(X))/(n(S)) = 16/36 = 4/9`

When X = 1, i.e. 5 or 6 appear on exactly one of the dice, then

X = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4)}

∴ n(X) = 16

∴ P(X = 1) = `(n(X))/(n(S)) = 16/36 = 4/9`

When X = 2, i.e. 5 or 6 appear on both of the dice, then

X = {(5, 5), (5, 6), (6, 5), (6, 6)}

∴ n(X) = 4

∴ P(X = 2) =`(n(X))/(n(S)) = 4/36 = 1/9`

∴ The required probability distribution is

0 1 2
P(X = x) `4/9` `4/9` `1/9`
shaalaa.com

Solution 2

Success is defined as a number greater than 4 appears on at least one die

Let X denote the number of successes.

∴ Possible values of X and 0, 1, 2.

Let P(getting a number greater than 4) = p

= `2/6`

= `1/3`

∴ q = 1 – p

= `1 - 1/3`

= `2/3`

∴ P(X = 0) = P(no success)

= qq

= q2

= `4/9`

P(X = 1) = P(one success)

= qp + pq = 2pq

=  `2 xx 1/3 xx 2/3`

= `4/9`

P(X = 2) = P(two successes)

= pp

= p2

= `1/9`

∴ Probability distribution of X is as follows:

0 1 2
P(X = x) `4/9` `4/9` `1/9`
shaalaa.com
  Is there an error in this question or solution?
Chapter 2.7: Probability Distributions - Short Answers II

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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) 0.1 0.6 0.3

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

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.


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


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)`.


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:

P.d.f. of a.c.r.v X is f (x) = 6x (1 − x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere)

If P (X < a) = P (X > a), then a = .....


If the p.d.f. of c.r.v. X is f(x) = `x^2/18`, for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) = ______. 


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) =


Solve the following :

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

The person on the high protein diet is interested gain of weight in a week.


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)


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)


Find the probability distribution of number of heads in four tosses of a coin


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).


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) = ______.


Choose the correct alternative :

X: is number obtained on upper most face when a fair die….thrown then E(X) = _______.


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) = _______


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 the distribution function of discrete r.v.x with p.m.f. P(x) = `(x - 1)/(3)` for x = 1, 2, 3 and P(x) = 0 otherwise then F(4) = _______.


Fill in the blank :

E(x) is considered to be _______ of the probability distribution of x.


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 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


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 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 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).


Solve the following problem :

Let X∼B(n,p) If E(X) = 5 and Var(X) = 2.5, find n and p.


If X denotes the number on the uppermost face of cubic die when it is tossed, then E(X) is ______


Choose the correct alternative:

f(x) is c.d.f. of discete 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) = ______


If p.m.f. of r.v. X is given below.

x 0 1 2
P(x) q2 2pq p2

then Var(x) = ______


If X is discrete random variable takes the values x1, x2, x3, … xn, then `sum_("i" = 1)^"n" "P"(x_"i")` = ______


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

k = `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
  1. Determine the value of k.
  2. Find P(X ≤ 4)
  3. P(2 < X < 4)
  4. P(X ≥ 3)

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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×