# MLE 2 coins, 6 flips

I am a bit confused on the way to go for this exercise could you please help me :

You have five coins in your pocket. It is known a priori that one coin has heads with probability 0.4 and the other four coins give heads with probability 0.7

A coin is being pulled out one of the five coins at random from your pocket (each coin has probability $$\frac{1}{5}$$ of being pulled out), and you want to find out which of the two types of coin it is. To that end, you flip the coin 6 times and record the results of each coin flip where $$X_1 = 1$$ if “heads" and $$X_1 = 0$$ if “tails".

Suppose that $$X_1 +X_2+X_3+X_4+X_5+X_6 = 3$$. Find the maximum likelihood estimate $$\hat{p}^{MLE}$$ for using the given data.(Enter a numerical value accurate to 3 decimal places.)

My first issue is that how can i find a single p mle when I have 2 probabilities for head. I proceeded as follows :

$$\prod_{i = 1}^{6} \left[\frac{1}{5} p^{x_i} (1-p)^{1-x_i} + \frac{4}{5} p^{x_i} (1-p)^{1-x_i}\right]$$

The $$x_i$$ will be replaced by 3, 1 will become $$n$$ i.e 6, which then becomes 3..

Then I will differentiate with respect to $$p$$ which will give me $$0.5$$ and this is wrong.

Thank you

SB

The probability of observing $$3$$ heads from $$6$$ flips is $$20 p^3 (1-p)^3$$. There are two possible values of $$p$$:

• when $$p=0.4$$ this is $$0.27648$$
• when $$p=0.7$$ this is $$0.18522$$

so the maximum likelihood estimate corresponds to the higher likelihood and would be $$\hat{p}=0.4$$

That calculation took no account of the fact that there were originally one coin with $$p=0.4$$ compared to four coins with $$p=0.7$$. If you had, then the more likely value for $$p$$ would have $$0.7$$, found by considering the posterior probability that the flipped coin had $$p=0.4$$, a Bayesian calculation you were not asked for.

If you had been, then the answer would be $$\dfrac{\frac15 0.27648}{\frac15 0.27648 + \frac45 0.18522}= \dfrac{128}{471} \approx 0.272$$, which is higher than the original $$\frac15$$ but lower than the posterior probability of about $$0.728$$ that the flipped coin had $$p=0.7$$

• in the question, it's been asked that the answer should be given with 3 decimal places, with 0.4 an answer, I tend to think that may be its not the correct one? May 4 '19 at 10:28
• Siting: There are only two possible reasonable answers: $p=0.4$ or $p=0.7.$
– whuber
May 4 '19 at 22:35