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Numerade Educator

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Problem 74 Hard Difficulty

Consider the following information about travelers (based partly on a recent Travelocity poll): 40$\%$ check work e-mail, 30$\%$ use a cell phone to stay connected to work, 25$\%$ bring a laptop with them, 23$\%$ both check work e-mail and use a cell phone to stay connected, and 51$\%$ neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. Finally, 88 out of every 100 who bring a laptop check work e-mail, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop.
(a) What is the probability that a randomly selected traveler who checks work e-mail also uses a cell phone to stay connected?
(b) What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected?
(c) If a randomly selected traveler checked work e-mail and brought a laptop, what is the probability that s/he uses a cell phone to stay connected?

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Video Transcript

okay for this problem. We have that 40% of people on vacation use their laptops. And actually I'll yeah, I'll leave that as Event A is using their sorry, 40% of people use their emails. You have 30% of people use their cell phones, labeling that as B and way have that 25% of people use their laptops, so that is events. See also, given that of the or other that there's 23% chance of somebody thio use their email actually labeled email. This is cell. This is laptop probability of somebody using their email and their laptop equal to 0.23 Probability of somebody not using any of the list of devices. A not and be not and see not or not, is equal to 0.51 of the probability that a given C is equal to 0.88 Probability of C given B is equal to 0.7. So for our first question here, A. We want the probability of be given a that's probability of using cell phone given that they used in their email that is equal to the probability of B and C divided by the probability of C which itself is equal to the probability. Oh, sorry, the B and C N N b and A. Yeah, which is are divided by probability of a, which is equal to the probability of A and B divided by the probability of a just changing it. So we have it in the exact form we had it up there. They're the exact same. That's 0.23 divided by 0.4 comes out to 0.575 part B want. The probability of be given see is equal to the probability of B and C divided by the probability of C equal to the probability of C given b times the probability of B divided by the probability of C where we know the probability of c given b so that is 0.7 times 0.3 divided by 0.25 which comes out to 0.84 for C. You have We want to find the probability of B given A and C so this one is a little bit trickier. What we want to do here This is actually where we use that knowledge that 51% of people don't use any of their devices. So that means that probability of somebody using A or B or C is equal to one minus 0.51 equals 0.49 So we also and this is in the textbook. It's in the believe it's in the axioms and interpretations section. Um, yeah, it's in the axioms and interpretation section of the textbook. In the version that I have here. It's about Paige, 37 or actually page 13. Rather, um, but we have that The probability of A or B or C is equal to the probability of a plus probability of B plus probability of C minus probability of a and B minus the probability of a and see minus the probability of B and see, plus the probability of A and B and C. So the reason why I'm going off on this is actually that probability of A and B and C right there, because probability of be given A and C is the same thing as the probability of A and B and see divided by the probability of I believe it should be it's divided by the probability of Excuse me. It is divided by the probability of A and C so we can use this formula down here to get the probability of A and B and C plug it in up there and we already have the probability or a probability of A and C is easy to get probability of A and C people, to the probability of a given see which we have and the probability of C, which we also have So rearranging this equation we end up getting that s o rearranging this equation and dividing by a given see times. PFC basically substituting in mhm rearranging this equation for p of A and B N c substituting into the p of be given A and C equation. If the probability of be given A and C is equal to 0.49 minus p of a minus p of B minus p of C plus p of and now with the A and C things. So I believe we have the probability of A and C no, we don't not explicitly given so probability of A and C we can write as the probability of I'm sorry, probably of A and B. We have So leave that in that form than probability of a and see we don't have. We can write that as probability of a given see times the probability of C the probability of a or Serie B N c then going to be, plus the probability of C given b times probability of being now. Last thing is that divide everything their i p of a given see as PSC so going through and substituting in all of our numbers here should get 0.49 minus 0.4, minus 0.3 minus 0.25 plus 0.23 I'm sorry. 0.23 plus 0.8 eight 10 0.25 plus 0.7 times 0.3 all divided by 0.88 I'm 0.25 which is going to equal once we go through and throw that into a calculator. 0.909