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प्रश्न
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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उत्तर १
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.
उत्तर २
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.
उत्तर ३
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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