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Multiple Choice
A commuter train arrives at a station once every 30 minutes. If a passenger arrives at the station at a random time, what is the probability that the passenger will wait less than 10 minutes?
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Verified step by step guidance
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Step 1: Recognize that this problem involves a uniform probability distribution, as the train arrives every 30 minutes and the passenger's arrival time is random. In a uniform distribution, the probability density function is constant over the interval of interest.
Step 2: Define the interval of the uniform distribution. Here, the random variable X (the waiting time) is uniformly distributed between 0 and 30 minutes. The probability density function is constant and can be calculated as \( f(x) = \frac{1}{b-a} \), where \( a \) and \( b \) are the endpoints of the interval.
Step 3: Calculate the probability that the passenger waits less than 10 minutes. This is equivalent to finding the area under the probability density function from \( x = 0 \) to \( x = 10 \). The formula for the probability is \( P(X < 10) = \int_{0}^{10} f(x) dx \).
Step 4: Substitute the value of \( f(x) \) into the integral. Since \( f(x) = \frac{1}{30} \), the integral becomes \( P(X < 10) = \int_{0}^{10} \frac{1}{30} dx \).
Step 5: Solve the integral. The result of the integral will give the probability that the passenger waits less than 10 minutes. Simplify the expression \( P(X < 10) = \frac{1}{30} \times (10 - 0) \).