91Ó°ÊÓ

Required probability will follow exponential distribution with decay rate \(\frac{1}{7}\) so the required probability can be given by,

\(P(x<2)=\left ( 1-e^{-\frac{1}{7}\times 2} \right )\)

\(P(x<2)=1-0.7515\)

\(P(x<2)=0.2485\)

Therefore, probability that the time between two successive visits to the urgent care facility is less than \(2\) minutes is \(P(x<2)=0.2485\).

Required probability can be calculated by,

\(P(x>15)=1-(x<15)\)

\(P(x>15)=1-\left ( 1-e^{-\frac{1}{7}\times 15} \right )\)

\(P(x>15)=e^{-\frac{1}{7}\times 15}\)

\(P(x>15)=0.1173\)

Therefore, the probability that the time between two successive visits to the urgent care facility is less than \(15\) minutes is \(P(x>15)=0.1173\).

The required probability can be written as \(P(x>15|x>10)\) using the memory less exponential distribution is given by,

\(P(x>t+r|x>t)=P(x>r)\)

\(P(x>t+r|x>t)=P(x>10+5|x>10)\)

\(P(x>t+r|x>t)=P(x>5)\)

\(P(x>t+r|x>t)=1-P(x\leq 5)\)

\(P(x>t+r|x>t)=1-\left ( 1-e^{-\frac{1}{7}\times 5} \right )\)

\(P(x>t+r|x>t)=e^{-\frac{1}{7}\times 5}\)

\(P(x>t+r|x>t)=0.4895\)

Therefore, the probability that the next person will arrive within the next five minutes is \(P(x>t+r|x>t)=0.4895\).