# Need Clarity in Finding Expectation of X, and Y

If there are four buses each bus has 40,33,25,50 students respectively. If X,Y are two random variables, X represents the no of students in the bus of the selected student , Y represents the no of students in the bus of a selected driver.

Ex: If student from bus1 is selected then X = 40 , Driver from bus2 is selected then Y=33.

The question is ,whether E(X) is greater than E(Y). Find E(X) and E(Y).

I thought of different approach of solving this question.That is suppose X=40 (means student selected from Bus1), Based on My Idea the $$P(x=40)= P(\text{Bus1} \mid \text{Student selected})$$ (Bayes rule probability , why because we know that the student is selected then the probability of the selected comes from the bus1 )

$$PX(x=40) = P(\text{Bus1} \mid \text{Student selected})$$

$$P(\text{Bus1} \mid \text{Student selected}) = \frac{P(\text{Bus1}) \cdot P(\text{Student selected} \mid \text{Bus1})}{P(\text{Student selected})}$$

Answer I got $$P(X) = \frac{\left( \frac{1}{4} \times 1 \right)}{\left( \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \right)}$$ , here 1/4 is probability of selecting a bus . Only students are there in the bus P(Student Selected / Bus1) = 1.

Why I thought Bayes rule is, In our sample space contains four buses with students,so selecting students associated with the sample space.And also why finding P(x=value) does not consider conditional probability.

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• I assume you mean all selections to be uniform, though that should be stated explicitly. $E(Y)$ is easy to compute, why not start there? $E(X)$ is not much harder...it's the same computation with uneven weightings.
– lulu
yesterday
• Not sure why you see this as a Bayes' problem. We aren't conditioning on anything. There are $4$ equally likely values $Y$ can take (assuming I'm right about uniformity). All you need to do is to average them.
– lulu
yesterday
• This reminds me of something a professor of mine told during prob theory class. It was a problem with the same principle as the one underlying in this question and at the end he explained it as: “ when you are randomly picking strands of spaghetti from a bag, the larger ones have a higher probability of being picked.” 19 hours ago
• Thank you for your details, mentioned here.It is helpful for me. 9 hours ago

here 1/4 is probability of selecting a bus .

You are not selecting from busses without bias. You are selecting from the students who travel on the four busses in a biased assortment of $$40$$, $$33$$, $$25$$, and $$50$$ students each bus.

The probability that a selected student is on a bus is equal to the count for students on that bus divided by the total count for students.

There are $$40$$ students who will be on the bus with $$40$$ students, and so:

$$\mathsf P(X=40)= \dfrac{40}{148}$$

Take a well shuffled deck of only $$12$$ hearts, $$10$$ diamonds, $$8$$ spades, and $$6$$ clubs, then pick a card. What is the probability that you picked a heart?

Is it (a) $$1/4$$ because there are four suits, or (b) $$1/3$$ because their is not a equal amount in each suit?

• Thanks for you reply , The probability of picking a heart from the well shuffled deck is 1/3 only( 12/36). Yes there should be some bias in picking up a student. Students from a bus with more number of students has more probability to select than other bus students. 9 hours ago
• Sorry to bother , I understood your explanation. I have a question, it is slightly a changed version for the above, suppose if we have three buses , bus1 contains 30 students and 10 drivers, bus2 = 20 students, 15 drivers, bus3 contains 10 students and 10 drivers. then what is the probability that X takes value 30. is it still 30/60. 9 hours ago
• Yes. If none of the students are also drivers, then $X$ and $Y$ will be independent. 4 hours ago

This answer provides an intuitive (somewhat non-math-oriented) explanation of what is happening.

For illustrative purposes, consider the alternate problem of 3 busses, with number of students on each bus respectively equal to 1, 2, and 3.

Then, the expected value of $$~Y~$$ (i.e. focusing on the bus driver) is $$~(2).~$$

For the expected value of $$~X,~$$ you have

$$\left[ ~1 \times \frac{1}{6} ~\right] + \left[ ~2 \times \frac{2}{6} ~\right] + \left[ ~3 \times \frac{3}{6} ~\right] = \frac{14}{6} > 2. \tag1$$

What is actually happening is that in (1) above, the scalar of $$~3~$$ is being given three times the weight of the scalar of $$~1.~$$ So, it is as if you are determining the average value in two different sets:

$$\{1,2,3\} ~~~~\text{versus} ~~~~\{1,2,2,3,3,3\}.$$

Since this attribute of the problem, giving higher weight to the larger scalars, will automatically persist in the general problem, it will also persist in the specific problem presented by the original poster.

That is, since the number of students on the various busses are distinct, higher weight will automatically be given to the scalars associated with busses that have a greater number of students.

In the specific problem presented by the original poster, the camouflauge is that the difference in the number of students on each bus is not that pronounced. Therefore, the difference in size of the scalars is more subtle.

• Ok. Now I am able understand a little bit, with your explanation. 9 hours ago