Question

One of your friends eats lunch in the cafeteria sometime between 12:00 P.M. And 12:30 P.M. Another friend eats lunch in the cafeteria sometime between 12:15 P.M. And 12:45 P.M. Today you get to the cafeteria at 12:20 P.M. What is the probability that you have missed both friends in the cafeteria?.

Answer 1

Answer:

11.11% probability that you have missed both friends in the cafeteria

Step-by-step explanation:

We use the uniform probability distribution to solve this question, since each lunch duration is equally as likely.

Uniform distribution:

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

Probability that you miss the first friend:

Somewhere between 12 and 12:30 pm, so his lunches are between 0 and 30 minutes, so $a = 0, b = 30$

You arrive at 12:20. So if his lunch is 20 or less minutes, you miss him.

$P_{1} = P(X \leq 20) = \frac{20 - 0}{30 - 0} = 0.6667$

Probability that you miss the second friend:

Somewhere between 12:15 and 12:45 pm, so his lunches are between 0 and 30 minutes, so $a = 0, b = 30$.

You arrive at 12:20. So if his lunch is 5 or less minutes, you miss him.

$P_{2} = P(X \leq 20) = \frac{5 - 0}{30 - 0} = 0.1667$

Probability that you miss both friends:

Missing the first friend and the second friend are independent events. So we multiply these probabilities.

$p = P_{1}*P_{2} = 0.6667*0.1667 = 0.1111$

11.11% probability that you have missed both friends in the cafeteria