91Ó°ÊÓ

It is given that$F\left(t\right)=1-e^{-0.1t}$. We need to determine how many minutes are needed for the probability to reach$50$%.

$50$% is $0.5$substitute $0.5$to $F\left(t\right)$in role="math" localid="1647185215992" $F\left(t\right)=1-e^{-0.1t}.$

$0.5=1-e^{-0.1t}.$

$0.5-1=-e^{-0.1t}.$

$e^{-0.1t}=0.5.$

$-0.1t=\ln \left(0.5\right).$

$t=\frac{\ln \left(0.5\right)}{0.1}.$

$t≈6.93$min.

It is given that$F\left(t\right)=1-e^{-0.1t}.$We need to determine how many minutes are needed for probability to reach$80$%.

$80\%$is $0.8$. Substitute $0.8$to $F\left(t\right)$in role="math" localid="1647185499742" $F\left(t\right)=1-e^{-0.1t}.$

$0.8=1-e^{-0.1t}.$

$0.8-1=-e^{-0.1t}.$

$-0.2=e^{-0.1t}.$

$t=\frac{\ln \left(0.2\right)}{-0.1}.$

$t≈16.09$min.

It is given that $F\left(t\right)=1-e^{-0.1t}.$We need to explain it is possible for the probability to equal$100\%.$

$100\%$is $1,$substitute $1$to $F\left(t\right)$in,

$F\left(t\right)=1-e^{-0.1t}.$

$1-=1-e^{-0.1t}.$

$-e^{-0.1t}=0.$

The value of$e^{-0.1t}$will never is zero. Thus it is not possible for the probability to be$100\%.$