Please solve this problem. I guess the answer is mle of theta = 50 choose 30 * samplea mean of x
No. Answer should not be mle theta = 50.
here you are considering distribution of red ball as binomial distribution that is whenever a ball is drawn a replacement of it is done. which is not the case in question. If replacement would be the case. than mean of sample should probability of red ball multiplies total sample size that is 1/2*30 = 15 and thus E[theta] = 1/2*50 = 25. Which is not the case here.
Since here ball is getting drawn without replacement, it is the case of hypergeometric distribution.
As an example, if you have one red ball and one blue ball and you are trying to take a random ball out of it. Probability is obviously 0.5 for the first trial. Now, if you have replacement, you are guaranteed to go back to the condition that you have two balls in each trial, so probability is the same for each trial=0.5. However, if I don't replace the first ball that I take, then my next pick will be 100% for the ball that is still left and 0% for the ball already taken. Hence trials are dependent.
Please solve this problem. I guess the answer is mle of theta = 50 choose 30...
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Abdul-Rahim Taysir
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