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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. - Mathematics and Statistics

Sum

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.

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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.

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Part I | Q 1.06 | Page 155
Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise | Q 6 | Page 242
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