Advertisements
Advertisements
Question
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.
Solution 1
When a fair coin is tossed 4 times then the sample space is
S = {HHHH,HHHT,HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n (S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4 When X = 0,
then X= {TTTT}
∴ n (X) = 1
∴ P (X=0) = `(n(x))/(n(s))= 1/16 = {""^4C_0}/16`
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n (X) = 4
∴ P (X=1) = `(n(x))/(n(s)) = 4/16 = {""^4C_1}/16`
When X = 2, then
X ={ HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n (X) = 6
∴ P (X=2) = `(n(x))/(n(s)) = 6/16 = {""^4C_2}/16`
When X = 3, then
X ={ HHHT, HHTH, HTHH, THHH}
∴ n (X) = 4
∴ P (X=3) = `(n(x))/(n(s))= 4/16 = {""^4C_3}/16`
When X = 4, then
X = {HHHH}
∴ n (X) = 1
∴ P (X=4) = `(n(x))/(n(s)) = 1/16 = {""^4C_4}/16`
∴ the probability distribution of X is as follows :
| x | 0 | 1 | |||
| p(x) | `1/16` | `4/16` | `6/16` | `4/16` | `1/16` |
Also, the formula for p.m.f. of X is
P (x) = `{""^4C_x}/16`
x = 0,1,2,3,4
= 0 otherwise.
Solution 2
When a fair coin is tossed 4 times then the sample space is
S = {HHHH,HHHT,HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n (S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4 When X = 0,
then X= {TTTT}
∴ n (X) = 1
∴ P (X=0) = `(n(x))/(n(s))= 1/16 = {""^4C_0}/16`
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n (X) = 4
∴ P (X=1) = `(n(x))/(n(s)) = 4/16 = {""^4C_1}/16`
When X = 2, then
X ={HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n (X) = 6
∴ P (X=2) = `(n(x))/(n(s)) = 6/16 = {""^4C_2}/16`
When X = 3, then
X ={ HHHT, HHTH, HTHH, THHH}
∴ n (X) = 4
∴ P (X=3) = `(n(x))/(n(s))= 4/16 = {""^4C_3}/16`
When X = 4, then
X = {HHHH}
∴ n (X) = 1
∴ P (X=4) = `(n(x))/(n(s)) = 1/16 = {""^4C_4}/16`
∴ the probability distribution of X is as follows :
| x | 0 | 1 | 2 | 3 | 4 |
| p(x) | `1/16` | `4/16` | `6/16` | `4/16` | `1/16` |
Also, the formula for p.m.f. of X is
P (x) = `{""^4C_x}/16`
x = 0,1,2,3,4
= 0 otherwise.
Solution 3
A coin is tossed 4 times.
∴ n(S) = 24 = 16
Let X be the number of heads.
Thus, X can take values 0, 1, 2, 3, 4
When X = 0, i.e., all tails {TTTT},
n(X) = `""^4"C"_0` = 1
∴ P(X = 0) = `(1)/(16)`
When X = 1, i.e., only one head.
n(X) = `""^4"C"_1` = 4
∴ P(X = 1) = `(4)/(16)`
When X = 2, i.e., two heads.
n(X) = `""^4"C"_2 = (4!)/(2!2!)` = 6
∴ P(X = 2) = `(6)/(16)`
When X = 3, i.e., three heads.
n(X) = `""^4"C"_3` = 4
∴ P(X = 3) = `(4)/(16) = (1)/(14)`
When X = 4, i.e., all heads ≅ {HHHH},
n(X) = `""^4"C"_4` = 1
∴ P(X = 4) = `(1)/(16)`
Then,
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | `(1)/(16)` | `(4)/(16)` | `(6)/(16)` | `(4)/(16)` | `(1)/(16)` |
∴ Formula for p.m.f. of X is
P(X) = `(((4),(x)))/(16), x` = 0, 1, 2, 3, 4
= 0, otherwise.
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) | 0.1 | 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 |
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
If a r.v. X has p.d.f.,
f (x) = `c /x` , for 1 < x < 3, c > 0, Find c, E(X) and Var (X).
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 =
Choose the correct option from the given alternative:
If the p.d.f of a.c.r.v. X is f (x) = 3 (1 − 2x2 ), for 0 < x < 1 and = 0, otherwise (elsewhere) then the c.d.f of X is F(x) =
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 = x) = `((c_(x)^5 ))/2^5` , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise If a = P (X ≤ 2) and b = P (X ≥ 3), then E (X ) =
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.
Amount of syrup prescribed by physician.
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 (–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)
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 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).
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 \[\sum\limits_{i=1}^{n}\text{P}(x_i)\] = _______
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 :
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 E(X) = 5 and Var(X) = 2.5, find n and p.
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) = ______
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 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 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 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]
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")` = ______
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 |
Complete the following activity.
Solution: Since `sum"p"_"i"` = 1
P(X ≥ 3) = `square - square - square = 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 following function represents the p.d.f of a.r.v. X
f(x) = `{{:((kx;, "for" 0 < x < 2, "then the value of K is ")),((0;, "otherwise")):}` ______
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)
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)
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)
