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We throw a die independently four times and let X denote the minimal value rolled. (a)...

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Basic Math

We throw a die independently four times and let X denote theminimal value rolled. (a) What is the probability that X ? 4. (b)Compute the PMF of X. (c) Determine the mean and variance of X.

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3.9 Ratings (413 Votes)

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
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