Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
In a made-up school, of the students come by bus. On a given day, a bus student is late with probability , and any other student is late with probability . Find the probability that a student picked at random is late, and the probability that the student does not come by bus and is late. Two students are then picked at random, and the school assumes that their days are independent. Find the probability that both are on time. Give four decimal places.
In a probability tree, multiply along a path and add the paths that make the event. Without replacement, the second branches use the counts that are left. From a big population the draws are almost independent.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Late. Bus and late: . Not bus and late: . Together: .
Step 2. Not bus and late. From Step 1, .
Step 3. Both on time. One student is on time with probability . Two independent students: .
Answer: , and about .
Watch out. If both students take the same bus, their days are not independent. One late bus makes both late.
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