Question

At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.

a. Find the probability that the time between two successive visits to the urgent care facility is less than \(2\) minutes.

b. Find the probability that the time between two successive visits to the urgent care facility is more than \(15\) minutes.

c. If \(10\) minutes have passed since the last arrival, what is the probability that the next person will arrive within the

next five minutes?

Short answer

Part a. \(P(x<2)=0.2485\)

Part b. \(P(x>15)=0.1173\)

Part c. \(P(x>15|x>10)=0.4895\)

Step-by-step solution

Part a. Step 1. Explanation

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

Part b. Step 1. Explanation

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

Part c. Step 1. Calculation

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