Probability of Stopping at 2nd and 6th Sets of Traffic Lights

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SUMMARY

The probability of a person stopping at the 2nd and 6th sets of traffic lights, while passing through a total of 6 sets, is calculated using the independent probabilities of stopping and not stopping. Given a stopping probability of 0.7 and a non-stopping probability of 0.3, the formula to determine this specific scenario is (0.3)^4 * (0.7)^2. This results in the combined probability of stopping at the designated sets while not stopping at the others.

PREREQUISITES
  • Understanding of basic probability concepts
  • Familiarity with independent events in probability theory
  • Knowledge of probability multiplication rules
  • Ability to perform calculations involving exponents
NEXT STEPS
  • Study the concept of independent events in probability theory
  • Learn about binomial probability distributions
  • Explore real-world applications of probability in traffic systems
  • Investigate advanced probability topics such as Markov chains
USEFUL FOR

Mathematicians, statisticians, students studying probability theory, and anyone interested in understanding traffic light systems and their implications in urban planning.

TomZiel
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If a person passes through 6 sets of independent traffic lights that have a probability of 0.7 to stop the person at any particular set.

Then how would you figure out the probability of that person stopping on the 2nd and 6th sets only?
 
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That looks straight forward. If the probability of being stopped is 0.7, then the probability of not being stopped is 0.3. Since these events are independent, the probability of them happening together is their product. Of 6 sets of lights the person is to be stopped at a particular 2 and not at the other 4. That is:
(0.3)4(0.7)2.
 

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