In this problem, we have a scenario where somebody is going home and there are two possible routes. The first route has to railroad crossings and the second has four crossings. The person will be laid if on route at least half of the railroad crossings on the route. Um, he gets stopped at due to a train crossing and the probability of getting stopped at any individual crossing is 0.1. So for party were asked which route should be taken either route one or two to minimize the probability of being late? Let's calculate the probabilities of being laid for each of the roots. So the probability of being late given that we take Route one, So Route one has four crossing. So we have to be stopped at at least two of them in order to be late to is at least half of them. Now this is equal to the probability that were stopped sequel to one, minus the probability that were stopped at either zero or one crossings, and this is equal to one minus. So the probability of being stopped at zero crossings is 0.1. Sorry, that's should be 0.9 to the exponents for 0.9 is the probability of not being stopped at any individual crossing. So to not be stopped at four, the probability is 0.92 exponents for plus the probability of being stopped at one. So we have to have three successes. So that's 30.19 exponents three and one where were stopped. So that 0.1 to explain it one. And then there are four times to arrange these four ways to arrange these. So we get a probability of 0.5 to 3. And now for the second route, the probability of being late given that we take this route is the probability of being stopped at one or both crossings. And then we have to subtract from this intersection. And that's because being stopped at one intersection and being stopped at two intersections are not mutually exclusive events. This equals 0.1 plus 0.1 minus 0.1 squared. It comes out to 0.19 And so the conclusion is to take the first route because the probability of being laid is lower. Now, for part B, we're told that he takes either route with equal probability. And if he is late, what is the probability that he took the first route? So what is the probability that the first rate was taken, given that he is late and here we can apply Bayes rule? So this is equal to the probability of the first rate and being laid divided by the probability of being laid. So the numerator is theirs. Probability of taking the first Rufus half times the probability of being late 0.5 to 3. That's the probability of being late, given that the first rate has taken and in the denominator, we use the the rule of the law of total probability to find the probability of being late and thats half times zero point 05 to 3, plus half times 0.19 and this comes out to 0.216 So, given that he is late, the probability that he took the first route is 0.216