The following problem is given accordingly:

The path used by a certain Canada Post mailman to deliver mail to the University of Guelph includes two intersections with traffic signals. Probability of the event that he will have to stop at the first intersection is 0.35; and the probability that he will need to stop at the second one is 0.65. Furthermore, probability of making a stop at at least one of the two intersections is 0.7. What is the possibility that:

a) He will stop at both intersections?
b) The second intersection given that he stopped at the first one? Is this different from 0.65?

In this problem stated that P(AUB) = 0.7. However, if the problem was independent, P(AUB) = 0.7725.

I argued the fact that in order for the probabilities to be independent, the driver's chances at stopping on the second intersection would have to be the same before and after he was stopped at the first intersection. Realistically speaking however, this would not be the case hence I assumed that they were dependent.

If the lights are independent, why so and is there a way to prove this mathematically?

  • $\begingroup$ By seeing the difference between $0.7$ and $0.7725$, you have proved that the two lights are dependent. This dependence will also help you answer the second part of (b). $\endgroup$
    – player3236
    Oct 16 '20 at 3:58
  • $\begingroup$ You are correct, the two events are indeed dependent, and your calculation shows that. (You should not make subjective arguments when you can determine this by using given probabilities.) However, you should be able to answer the two questions regardless of dependence/independence. $\endgroup$
    – A.J.
    Oct 16 '20 at 3:59


$$\mathbb{P}[A \cap B]=0.35+0.65-0.7=0.3$$


$$\mathbb{P}[B|A]=\frac{0.3}{0.35}=\frac{6}{7} \ne 0.65$$

This result again shows you the events are not independent.


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