Riya leaves for school between 7:00 AM and 7:15 AM. She takes between 20 and 30 minutes to reach the school. Let 𝑋X denote her time of departure, and let 𝑌Y denote her travel time. Assuming that 𝑋X and 𝑌Y are continuous, independent and uniformly distributed, find the probability that she reaches the school before 7:30 AM. Enter the answer correct to two decimal places.
Question
Riya leaves for school between 7:00 AM and 7:15 AM. She takes between 20 and 30 minutes to reach the school. Let 𝑋X denote her time of departure, and let 𝑌Y denote her travel time. Assuming that 𝑋X and 𝑌Y are continuous, independent and uniformly distributed, find the probability that she reaches the school before 7:30 AM. Enter the answer correct to two decimal places.
Solution
To solve this problem, we need to consider two scenarios:
- Riya leaves at 7:00 AM and reaches school in 30 minutes, i.e., at 7:30 AM.
- Riya leaves at 7:15 AM and reaches school in 15 minutes, i.e., at 7:30 AM.
Since X and Y are independent and uniformly distributed, the joint probability density function (pdf) is given by:
f(x, y) = 1/(b-a) * 1/(d-c) = 1/15 * 1/10 = 1/150 for 0 <= x <= 15 and 20 <= y <= 30.
We need to find the probability P(X + Y < 30). This can be found by integrating the joint pdf over the region defined by X + Y < 30.
The limits of the integration are:
For X: 0 to 15 For Y: 20 to 30-X
The integral is:
∫ from 0 to 15 ∫ from 20 to 30-x (1/150) dy dx = ∫ from 0 to 15 [(30-x-20)/150] dx = ∫ from 0 to 15 [10-x)/150] dx.
Solving this integral gives us the probability that Riya reaches school before 7:30 AM.
The result of the integral is 0.3333 or 33.33% when rounded to two decimal places. So, the probability that Riya reaches school before 7:30 AM is 0.33 or 33.33%.
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