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

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

Solutions

Expert Solution

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

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

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) =μ=	ΣxP(x) =	1.7554
	E(x²) =	Σx²P(x) =	3.9915
	Var(x)=σ² =	E(x²)-(E(x))²=	0.9101
	std deviation=	σ= √σ² =	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) =μ=	ΣxP(x) =	1.7554
	E(x²) =	Σx²P(x) =	3.9915
	Var(x)=σ² =	E(x²)-(E(x))²=	0.9101

from above mean=1.7554

Var(X)=0.9101

milcah answered 3 months ago

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