This question from my textbook

Three high schools have senior classes of size 100, 400, 500. Scheme A: make a list of all 1000 students and choose one randomly; Scheme B: pick a school at random then a student at random;

Makes the comparison by showing that choosing a student from one of the high-schools is different across the two schemes

In Scheme A each person in first school is chosen with probability 1/1000; in Scheme B choose that school 1/3 of the time, and then each person is chosen 1/100 of the time, so a person in the first school is now chosen 1/300 of the time.

I did not think to use a single school as an example and wanted to solve this question by showing that in general the probability of choosing any student from any school is different.

How would I calculate the probability of choosing any particular student in scheme B?

I tried drawing a tree diagram where I had $\frac{1}{3}$ probability for choosing each school and then $\frac{1}{100}$, $\frac{1}{400}$, $\frac{1}{500}$ for choosing a student from each school respectively.

Is this a correct start?


Yes, that is a correct start. The book answer finishes for students of school A, now you just do the same for schools B and C.

  • $\begingroup$ So the probability is then the sum of probabilitoes for each school? $\frac{1}{300}+\frac{1}{1200}+\frac{1}{1500} = \frac{29}{6000}$ $\endgroup$ – EvaD Oct 5 '17 at 13:14
  • $\begingroup$ The probability of what? The probability that some student is chosen has to be $1$. This is a good check of your arithmetic. In fact $100 \cdot \frac 1{300}+400 \cdot \frac 1{1200}+500 \cdot \frac 1{1500}=1$. What you have shown is that the chance of an individual student from C being selected if $\frac 15$ of the chance that an individual student from A is selected That is what you were looking to show. $\endgroup$ – Ross Millikan Oct 5 '17 at 14:26
  • $\begingroup$ I am trying to find the probability of any given student being selected using scheme B. By scheme A $P(student) = \frac{1}{1000}$, by scheme B I am asking if it is correct that $P(student) = P(student | school A) + P(student| school B) + P(student| school C)$ $\endgroup$ – EvaD Oct 5 '17 at 14:39
  • $\begingroup$ The point is that there is not a single probability for any given student to be selected for scheme B. There is a different one for students of each school. You can certainly add the numbers like you did, but it doesn't mean anything. You can multiply them, too. The point is simply that under scheme B different students have different probabilities to be selected. $\endgroup$ – Ross Millikan Oct 5 '17 at 14:47

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