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There are $12$ blue marbles and $4$ red marbles in a bag. You reach into the bag and pull $5$ marbles out at random. What is the expected value of the number of blue marbles drawn?

I found the solution to this problem with a much simpler solution than the one that was offered.

The way I solved it: We have 5 marbles, and the probability of getting a blue marble is $12/16 = 3/4$.

Therefore the Expected Value is $5\cdot 3/4 = 15/4$.

I got the answer right but I'm not sure if I got it by fluke.

The way they solved it was by solving $P(X=1), P(X=2), ..., P(X=5)$ individually using combinations and summing these probabilities.

Seems a bit complicated to me if I can get the same answer the way I did it, unless my reasoning isn't sound.

Any thoughts?

callculus42
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dbqp
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1 Answers1

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There are a number of approaches to problems like these.

One such approach is by approaching directly by the definition, calculating the probability of each, etc...

The way that you used, perhaps unintentionally without understanding why it works, is via linearity of expectation. $E[\alpha X+\beta Y] = \alpha E[X]+\beta E[Y]$.

Here, we can describe the result of your problem with marbles as a random variable $X$ which counts the total number of blue marbles that you drew and we can temporarily assume that we drew out the marbles in sequence to make things easier rather than all at once. We can break this down as $X=X_1+X_2+\dots+X_5$ where $X_i$ is a random variable which takes value $1$ in the case that the $i$'th marble drawn is blue and $0$ otherwise.

Now, $E[X]=E[X_1+X_2+\dots+X_5]=E[X_1]+E[X_2]+\dots+E[X_5]=Pr(X_1=1)+Pr(X_2=1)+\dots+Pr(X_5=1)$

Now, noticing that $Pr(X_i=1)=\frac{12}{16}$ for all values of $i$, we get $E[X]=5\cdot \frac{12}{16}$ as you had calculated.

Notice that the probability that the $i$'th marble is blue is $\frac{12}{16}$ regardless of whether we were drawing with or without replacement.

JMoravitz
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  • Thanks for the answer. I just found it strange that in the solution they picked the marbles out one at a time and recalculated the probability each time using (relatively) complicated combinatorics. Why do that when you can pick out all the marbles at once with a simple probability. I guess it's just because they want to use the definition of expected value as that is what they are teaching. – dbqp Oct 07 '19 at 12:39