A die is thrown until the first time the total sum of the face values of the die is 700 or greater. Estimate the probability that, for this to happen,
Solution:
First lets calculate the total sum of the face values of the die.
\({ S }_{ n }={ X }_{ 1 }+{ X }_{ 2 }+...+{ X }_{ n }\)
The expected value and variance would be:
\(E(X)=\frac { 7 }{ 2 }\) and \(E({ S }_{ n })=n*\frac { 7 }{ 2 }\)
$V(X)= $ and \(V({ S }_{ n })=n*\frac { 35 }{ 12 }\) which means that \(σ({ S }_{ n })=\sqrt { n*\frac { 35 }{ 12 } }\)
\(P({ S }_{ 210 }<700)\)
\(E({ S }_{ 210 })=210*\frac { 7 }{ 2 }=735\)
\(σ({ S }_{ 210 })=\sqrt { 210*\frac { 35 }{ 12 } }=24.75\)
round((700 - 735) / 24.75, 4)## [1] -1.4141round(pnorm(-1.4141), 4)## [1] 0.0787\(P({ S }_{ 210 }<700)=0.0787\)
The probability that the total sum of the face values of the die requires more than 210 tosses is 0.0787.
\(P({ S }_{ 190 }\ge 700)\)
\(E({ S }_{ 190 })=190*\frac { 7 }{ 2 }=665\)
\(σ({ S }_{ 190 })=\sqrt { 190*\frac { 35 }{ 12 } }=23.54\)
round((700 - 665) / 23.54, 4)## [1] 1.4868round(1 - pnorm(1.4868), 4)## [1] 0.0685\(P({ S }_{ 190 }\ge 700)\)
The probability that the total sum of the face values of the die requires less than 190 tosses is 0.0685.
\(P({ S }_{ 180 }<700, P({ S }_{ 210 }\ge 700)\) =
\(P({ S }_{ 180 }<700)-P({ S }_{ 180 }<700, P({ S }_{ 210 }\ge 700)\) =
\(P({ S }_{ 180 }<700)-P({ S }_{ 210 }\ge 700)\)
\(P({ S }_{ 180 }<700)\)
\(E({ S }_{ 180 })=180*\frac { 7 }{ 2 }=630\)
\(σ({ S }_{ 180 })=\sqrt { 180*\frac { 35 }{ 12 } }=22.91\)
pnorm(22.91)## [1] 11 - 0.0757## [1] 0.9243The probability that the total sum of the face values of the die requires between 180 and 210 tosses is 0.0685.