Question

We roll a die until the sum of points obtained exceeds 350. Approximate the probability that we will roll more than 120 times.

Answer 1

here for a single roll probability of an outcome is 1/6

x	P(x)	xP(x)	x2P(x)
1	1/6	0.167	0.167
2	1/6	0.333	0.667
3	1/6	0.500	1.500
4	1/6	0.667	2.667
5	1/6	0.833	4.167
6	1/6	1.000	6.000
	total	3.500	15.167
	E(x) =μ=	ΣxP(x) =	3.5000
	E(x2) =	Σx2P(x) =	15.1667
	Var(x)=σ2 =	E(x2)-(E(x))2=	2.917
	std deviation=	σ= √σ2 =	1.7078

hence for 120 rolls; expected sum of numbers =120*3.5=420

and std deviation =1.7078*sqrt(120)=18.71

hence probability that we will roll more than 120 times untill sum exceed 350

=P(in 120 rolls sum reached at most 350)=P(X<=350)=P(Z<(350.5-420)/18.71)=P(Z<-3.71)

=0.0001