# Thread: Max likely hood estimator problem in Statistics.

1. ## Max likely hood estimator problem in Statistics.

I have tried to solve the problem, but I got the max likely hood estimate for p is 2/3, but it is given either 1/ 3 or 1/2. So please clear out my confusion.

The number of red chips and white chips in an urn is unknown, but the proportion, p, of reds is either 1/ 3 or 1/2 . A sample of size 5, drawn with replacement, yields the
sequence red, white, white, red, and white. What is the maximum likelihood estimate for p?

2. ## Re: Max likely hood estimator problem in Statistics.

If the proportion is $p$, sampling with replacement gives the probability of a sample of the sequence $\rm{rwwrw}$ is:

$$Prb({\rm{rwwrw}}|p)=p(1-p)(1-p)p(1-p)=p^2(1-p)^3$$

The likelihood of $p$ is then:

$$Lik(p|{\rm{rwwrw}})=Prb({\rm{rwwrw}}|p)=p^2(1-p)^3$$

The two likelihoods for $p=1/2$ and $p=1/3$ are $\sim 0.03125,\ \sim 0.03292$ so the latter is the greater. Hence $1/3$ is the maximum likelihood estimator for $p$

If we remove the constraint that $p=1/2$ or $p=1/3$ but allow $p$ to take any value in $[0,1]$. Then the maximum likelihood estimator for $p$ is $0.4$ (which is closer to $1/3$ than $1/2$).

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3. ## Re: Max likely hood estimator problem in Statistics.

Thanks,

I am trying to solve the problem for several days, but for every efforts of mine end up with results nearby 18, so please guide me about how to solve the problem!!

Josh takes a twenty-question multiple-choice exam where each question has five possible answers. Some of the answers he knows, while others he gets right just by
making lucky guesses. Suppose that the conditional probability of his knowing the answer to a randomly selected question given that he got it right is 0.92. How many of the
twenty questions was he prepared for?

4. ## Re: Max likely hood estimator problem in Statistics.

Thanks,

I am trying to solve the problem for several days, but for every efforts of mine end up with results nearby 18, so please guide me about how to solve the problem!!

Josh takes a twenty-question multiple-choice exam where each question has five possible answers. Some of the answers he knows, while others he gets right just by
making lucky guesses. Suppose that the conditional probability of his knowing the answer to a randomly selected question given that he got it right is 0.92. How many of the
twenty questions was he prepared for?
First we need Bayes' rule: $p(k|r)p(r)=p(r|k)p(k)$, where $k$ represents knows and $r$ represents right. We are given $p(k|r)=0.92$, we may assume $p(r|k)=1$, and that $p(r)=p(r|k)p(k)+p(r|\lnot k)(1-p(k))$ and if he does not know and choses an answer at random we have $p(r|\lnot k)=0.2$. Hence $p(r)=0.8p(k)+0.2$

Now that Josh is prepared for a question means that he knows the answer and vice-versa, so we need to find $$p(k)=\frac{p(k|r)p(r)}{p(r|k)} =\frac{0.92 \left[0.8p(k)+0.2 \right]}{1} =0.92[0.8p(k)+0.2]$$

Now you solve this for $p(k)$ which gives the proportion that he knows the answer to, and multiply this by $20$ to get the number he is prepared for (which is about $13.9$ which should be rounded to $14$ )