65 blue marbles are placed into a bag with red marbles in it. Then 32 marbles (27 of them were red) were drawn out of the bag.Expected number of red…

Question

65 blue marbles are placed into a bag with red marbles in it. Then a sample of 32 marbles were drawn out of the bag. It was noted that of the 32 drawn out, 27 of them were red. What is the expected number of red marbles in the bag to begin with.

Answer 1

Since we have drawn out 5 blues and 27 reds then the expected is that it's a proportion of blues and reds in the box. Since $\frac{65}{5}=13$ if we denote by $x$ the number of reds in the box we should have $\frac{x}{27}=13$ which leads to $x=351$. Then we expect 351 red marbles at beginning.

Answer 2

The question regarding the expected value of the red marbles initially in the bag cannot be answered without an assumption about their distribution. However if we assume that there are at least $27$ red marbles then the most probable number of red marbles given the result is independent of this distribution.

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Such an assumption could be that the possible number of the red marbles is either $27$ or $28$ or ... $27+n$ and these possibilities are of equal probability, that is, $$P(R=i)=\frac1n, \ i=27,28,\dots, 27+n.$$ Based on this assumption, the expected value of the red marbles in the bag, $$E[R]=\frac1{100}[27+28+\cdots27+n]=\frac12[54+n].$$ Also, we have to assume that the drawing is independent from $R$. Having said that, we have $R+65$ marbles in the bag and we draw $32$ marbles and we find $27$ red ones in the sample. If $R_n$ denotes the number of the red marbles in the sample then we may be interested in $$P(R_n=27\mid R=n)=\frac{n\choose 27}{n+65\choose 32}$$ because it would be a sound choice to find an $n$ that maximizes the probability above -- according to the maximum likelihood principle.

The maximum occurs about $n=350$. So, the expectation is $\approx 202$.

The exact value based on this maximum likelihood estimation is $351$ as it was calculated by Mostafa. (See the following graph

here the figures on the vertical axis have to be multiplied by $10^{-8}$.)

Answer 3

If the number of red balls in a $32$-balls drawing is 27, then we assume that the probabilty of a ball in the bag to be red is $p=\frac{27}{32}$.

If there were $N$ balls in the bag, the expected number of red balls is $Np=\frac{27}{32}N$. But, there are $N-65$ red balls. So, $\frac{27}{32}N=N-65$ and $N=416$ is the expected number of balls in the bag, and $416-65=351$ is the expected number of red balls in the bag.