total number of outcomes =64 =1296 (As there are 4
outcomes possible on each roll)
a)P(X>=4)=P(X=4)+P(X=5)+P(X=6)=65/1296+15/1296+1/1296=81/1296
b)
Pmf of X:
P(X=1)=P(at least one dice show 1)=1-P(no dice show
1)=1-(5/6)4 =671/1296
P(X=2)=P(one dice from 6 have 2 and other three have 2 or
more)
=P(all dice shows 2 or more-all dice show 3 or
more)=(54-44)/64 =369/1296
P(X=3)=(44-34)/64 =175/1296
P(X=4)=(34-24)/64 =65/1296
P(X=5)=(24-14)/64 =15/1296
P(X=6)=1/1296
c)
| x |
f(x) |
xP(x) |
x2P(x) |
| 1 |
0.51775 |
0.518 |
0.518 |
| 2 |
0.28472 |
0.569 |
1.139 |
| 3 |
0.13503 |
0.405 |
1.215 |
| 4 |
0.05015 |
0.201 |
0.802 |
| 5 |
0.01157 |
0.058 |
0.289 |
| 6 |
0.00077 |
0.005 |
0.028 |
|
total |
1.755 |
3.992 |
|
|
|
|
|
E(x) =?= |
?xP(x) = |
1.7554 |
|
E(x2) = |
?x2P(x) = |
3.9915 |
|
Var(x)=?2 = |
E(x2)-(E(x))2= |
0.9101 |
|
std deviation= |
?= ??2 = |
0.9540 |
| x |
f(x) |
xP(x) |
x2P(x) |
| 1 |
671/1296 |
0.518 |
0.518 |
| 2 |
369/1296 |
0.569 |
1.139 |
| 3 |
175/1296 |
0.405 |
1.215 |
| 4 |
65/1296 |
0.201 |
0.802 |
| 5 |
15/1296 |
0.058 |
0.289 |
| 6 |
1/1296 |
0.005 |
0.028 |
|
total |
1.755 |
3.992 |
|
|
|
|
|
E(x) =?= |
?xP(x) = |
1.7554 |
|
E(x2) = |
?x2P(x) = |
3.9915 |
|
Var(x)=?2 = |
E(x2)-(E(x))2= |
0.9101 |
|
|
|
|
from above mean=1.7554
Var(X)=0.9101
