Question

You take a bus from your neighborhood to your school. The express bus arrives at your neighborhood at a random time between 7:30 and 7:36 a.m. The local bus arrives at your neighborhood at a random time between 7:30 and 7:40 a.m. You arrive at the bus stop at 7:33 a.m. Find the probability that you missed both the express bus and the local bus. Express your answer as a fraction in simplest form.

Answer 1

Answer:

$\frac{3}{20}$ probability that you missed both the express bus and the local bus.

Step-by-step explanation:

Uniform distribution:

An uniform distribution has two bounds, a and b.

The probability of getting a value lower than x is:

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

The express bus arrives at your neighborhood at a random time between 7:30 and 7:36 a.m.

6 minutes, so $a = 0, b = 6$

You arrive at 7:33, so if x < 3, you lose the express bus:

$P_E = P(X < 3) = \frac{3 - 0}{6 - 0} = \frac{3}{6} = \frac{1}{2}$

The local bus arrives at your neighborhood at a random time between 7:30 and 7:40 a.m.

10 minutes, so $a = 0, b = 10$

$P_B = P(X < 3) = \frac{3 - 0}{10 - 0} = \frac{3}{10}$

Find the probability that you missed both the express bus and the local bus.

$p = P_E*P_B = \frac{1}{2}*\frac{3}{10} = \frac{3}{20}$

$\frac{3}{20}$ probability that you missed both the express bus and the local bus.

Answer 2

The probability that you missed both the express bus and the local bus is 3/20

The individual probability of a uniform distribution is calculated as:

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

For the express bus

At 7:30 am, a = 0

At 7:36 am, b = 6

You arrived at 7:33 am; this means that:

x = 3

For you to miss the express bus, it means that the express bus left before 7:33 am.

i.e. x < 3

So, the probability that you miss the express bus is:

$P(X < 3) = \frac{3 - 0}{6 - 0}$

$P(X < 3) = \frac{3}{6}$

$P(X_E < 3) = \frac{1}{2}$

For the local bus

At 7:30 am, a = 0

At 7:40 am, b = 10

You arrived at 7:33 am; this means that:

x = 3

For you to miss the local bus, it means that the local bus left before 7:33 am.

i.e. x < 3

So, the probability that you miss the local bus is:

$P(X < 3) = \frac{3 - 0}{10 - 0}$

$P(X_L < 3) = \frac{3}{10}$

So, the probability that you miss both bus is:

$P(Both) = P(X_E < 3) * P(X_L < 3)$

This gives

$P(Both) = \frac 12 * \frac 3{10}$

$P(Both) = \frac 3{20}$

Hence, the probability that you missed both the express bus and the local bus is 3/20

Read more about probability at:

https://brainly.com/question/25870256