SOLUTION: Hello everyone, I'm stuck with this task..
A pedestrian can cross the street in 3 seconds. As he prepares to cross the street at a pedestrian crossing, the cars do not stop and do
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Question 1192267: Hello everyone, I'm stuck with this task..
A pedestrian can cross the street in 3 seconds. As he prepares to cross the street at a pedestrian crossing, the cars do not stop and do not miss him. The pedestrian starts walking only after looking around and making sure that no car will pass through him in 3 seconds. In this task we divide time in seconds. The probability that the car will pass in another second is 0.3. What is the probability that as long as a pedestrian can cross the street safely, only one car will pass by? Exactly two cars?
I'm appreciate for every answer!
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem:
**1. Define "Safe Crossing":**
A safe crossing means no car passes during a 3-second interval.
**2. Probability of No Car Passing in One Second:**
* The probability a car *will* pass in one second is 0.3.
* Therefore, the probability a car will *not* pass in one second is 1 - 0.3 = 0.7.
**3. Probability of No Car Passing in Three Seconds:**
* The probability of no car passing in three consecutive seconds is 0.7 * 0.7 * 0.7 = 0.7³ = 0.343. This is the probability the pedestrian can cross safely.
**4. Probability of Exactly One Car Passing During a Safe Crossing Period:**
This is a bit trickier. It means one car passes in the *total* time, but *not* during the 3-second crossing window.
* **Probability of one car passing in any given second:** 0.3
* **Probability of no car passing in any other second:** 0.7
* **Probability of one car passing in a specific second and no car passing in the other two seconds of the total period:** 0.3 * 0.7 * 0.7 = 0.147
* **Probability of one car passing in *any* of the three seconds of the total time and no car passing in the other two seconds of the total time:** 3 * 0.147 = 0.441 (We multiply by 3 because the car could pass in the first, second, or third second)
* Since the safe crossing window is 3 seconds, we need to consider cases where the one car passes *outside* of this window. To calculate this we need to know the total time period we are considering. Let's assume it is 4 seconds. Then there is only one second outside the 3-second crossing window. The probability of one car passing in this one second is 0.3. The probability of no car passing in the other 3 seconds is 0.7³ = 0.343. The probability of one car passing during the total 4 seconds is 0.3 * 0.7³ = 0.1029.
**5. Probability of Exactly Two Cars Passing During a Safe Crossing Period:**
Again, this means two cars pass in the total time, but *not* during the 3-second crossing window. Let's assume the total time is 5 seconds.
* **Probability of two cars passing in any two specific seconds and no car passing in the other three seconds:** 0.3 * 0.3 * 0.7 * 0.7 * 0.7 = 0.03087
* **Number of ways to choose two seconds out of five:** 5C2 = 10
* **Probability of exactly two cars passing in the 5 seconds:** 10 * 0.03087 = 0.3087
* We need two cars to pass in the two seconds outside the 3-second crossing window.
* The probability of this is 0.3 * 0.3 * 0.7³ = 0.03087
**Important Note:** The question is a bit ambiguous about the total time period we're considering. The calculations above assume a total time of 4 seconds for the "one car" scenario and 5 seconds for the "two car" scenario. If the total time period is different, the probabilities will change. You need to know the total time period for each scenario to calculate the correct probabilities.
Answer by ikleyn(52805) (Show Source): You can put this solution on YOUR website!
.
As the problem is formulated in the post, it is soup of words with an unknown meaning, but not a Math problem.
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