Conditional probability hard problem

Question

Suppose that we have three coins. One has two heads, while the others are standard. Two coins are flipped. In each case, the result is heads.

1. If you pick one of these two coins randomly, what is the probability that you will get the two headed coin?
2. If you pick the unflipped coin, what is the probability that you will get the two headed coin?

My attempt at 1:

Let A be the event: "We obtain two heads" and let B be the event: "We got the two-headed coin." We are looking for $$P(B|A)=\frac{P(B\bigcap A)}{P(A)}$$ Now, from Baye's Theorem: $$P(B\bigcap A)=P(A|B)P(B)=(\frac{1}{2})(\frac{1}{3})=\frac{1}{6}$$ I still need to find $P(A)$, could anyone help?

Answer 1

For the first question, ask yourself what's the probability that one of the coins picked are two-headed (event $A$) given that we obtained $2$ heads (event $B$)? There are two competing statements here:

$1-$Both of the coins are fair coins.
$2-$ One of the coins is a fair coin.

According to Bayes theorem, we have:

$P(A| B) = \frac{P(B| A) P(A)}{ P(B| A) P(A) + P(B| \neg A) P(\neg A)}$

Here we have $P(B|A)=\frac{1}{2}$ since flipping the coins are independent and the probability of getting head with the two-headed coin is $1$ while the probability of getting head with the other coin is $\frac{1}{2}$, we multiply these two.

We have $P(A)=\frac{\binom{1}{3}}{\binom{2}{3}}$ since we are picking two coins out of three in total and we need one of them to be the two-headed coin. This gives us that $P(\neg A)=1- \frac{\binom{3}{1}}{\binom{3}{2}}$

$P(B| \neg A)=\frac{1}{4}$ Since event $A$ not happening means that both of the coins picked are fair coins where the probability of getting head in each one of them is $\frac{1}{2}$ so we multiply this probability by itself then get $\frac{1}{4}$

I will leave the arithmetic to you, however, this gives us the probability that one of the coins picked is the two headed coin, then we need to multiply this probability by $\frac{1}{2}$ since we are picking on these $2$ coins at random and we need the probability we the two headed one.

Answer 2

When calculating $P(B\bigcap A)$, I think you made a mistake: $P(B) = 2/3$, not $1/3$.

Sorry I think you are right: $P(B) = 1/3$. I misunderstood what $B$ is. (I thought it is the event "the two headed coin is among the two flipped", but it is the event "you picked up the two headed coin", which has probability 1/3.)

To find P(A), use the total probability theorem: $P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$

Answer 3

Let's conduct $12$ trials. We'll pick each of the three possible pairs of coins in $4$ of those trials. Then we'll flip both of the coins we picked.

We'll get a pair of heads once out of the four trials where we picked two fair coins. The other two pairs of coins will each give us a pair of heads twice.

That means we end up with $5$ trials where we flipped two heads. In $1$ of those $5$ trials, we picked two fair coins. In the other $4$, one of our coins is the two-headed coin. That means there is a $\frac 45$ probability that one of our coins is two-headed so our chance of picking the two-headed coin if we randomly select one of the two flipped coins is $\frac 25 (= \frac 45 \cdot \frac 12)$ and the chance that the unflipped coin is the two-headed coin is $\frac 15$.