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. - Mathematics and Statistics

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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.

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