0
$\begingroup$

I had this question as a bonus problem on a previous exam and thought it was interesting, but I had no idea how to tackle it.

There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75% of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the first coin?

$\endgroup$
  • 1
    $\begingroup$ This is a problem for Bayes' theorem. No more no less. $\endgroup$ – Makina Dec 7 '18 at 0:46
0
$\begingroup$

One possibility to think about it "in a very naive way" is as follows. The coins are numbered as 1, 2, 3 - say. Instead of the given experiment, we can imagine an equivalent one. Consider a hat with $12$ tickets in it, labeled as $$ 1H,\ 1H,\ 1H,\ 1H;\ 2H,\ 2H,\ 2T,\ 2T;\ 3H,\ 3H,\ 3H,\ 3T. $$ Choosing a coin $k$ has the same probability as choosing a ticket $k?$ from the hat. For each fixed $k$, getting head (H) for the coin $k$ has the same probability as picking the ticket $kH$.

Now we pass to the conditional probability. We have only

$$\color{red}{1H,\ 1H,\ 1H,\ 1H};\ 2H,\ 2H;\ 3H,\ 3H,\ 3H $$ at our disposal. Picking $1H$ has then probability $$\frac 49\ .$$

$\endgroup$
0
$\begingroup$

Bayes' theorem approach:

$P(A_1), P(A_2)$ and $P(A_3) = \frac{1}{3}$ - choice of coins

$P(B)$ = get heads

You are asked a question to find $P(A_1 | B)$

$P(A_1 | B) = \frac{P(A_1)*P(B | A_1)}{P(B)} = \frac{P(A_1)*P(B | A_1)}{P(B | A_1)*P(A_1) + P(B | A_2)*P(A_2) + P(B | A_3)*P(A_3)} = \frac{4}{9} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.