I have a fair coin and a weighted coin which lands heads 75% of the time. I pick a coin at random and flip it 5 times and get heads 4 times and tails once. What is the probability that I picked the weighted coin?
Not sure on how to apply Bayes theorem to this question. Any help would be appreciated.
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$\begingroup$Hint: Let $W$ be the event of choosing the weighted coin, and $F$ be the event of flipping the coin you chose 5 times and getting 4 heads.
- You know that $P(W)=0.5$.
- You can calculate the probability that the weighted coin would give you 4 out of 5 heads -- this is $P(F\mid W)$.
- You can calculate the probability that the fair coin would give you 4 out of 5 heads -- this is $P(F\mid\neg W)$.
- With all that information, you can use the law of total probability to calculate $P(F)$
- finally, use Bayes' theorem to calculate $P(W\mid F)$ from the above, which is what you were asked for.
Well, you should indeed use Bayes' theorem, so you got that right!
Now, first calculate the probability of getting 4 heads with the fair coin ... let that be $P(4|F)$
Next calculate the probability of getting 4 heads with the biased coin ... let that be $P(4|B)$
Now, the probability of getting $P(4)$ at all is: $P(4|F)\cdot P(F) + P(4|B)\cdot P(B)$ ... and we know that $P(F)=P(B)=\frac{1}{2}$
Finally, we use Bayes' theorem:
$$P(B|4)=\frac{P(4|B)\cdot P(B)}{P(4)}$$
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