I came across a question that baffled me, and I'm not sure how to approach this
Suppose there are 6 coins: 3 coins have two Tails, 2 coins have two Heads, 1 is fair coin
All 6 are flipped at the same time, and before you can see the outcome of each, 5 of these coins are hidden away, and one is out in the open for you to see, and it has landed Heads.
What is the probability that the fair coin has landed Heads?
Now, this was an MCQ, and the minimum answer available was $\frac{3}{5}$. Also note, $\frac{2}{3}$ was not an option. I thought the events would be independent, how would knowing what the open coin's result is help us figure out what the result of the fair coin is? shouldn't it be $0.5$ either way? I may be naive, but I just can't get past this mental block 🙁
Edit:
All the answers are great, and I wish I could accept them all, thank you so much for your time!
Best Answer
Let's denote the event in which the fair coin lands on heads $A$, and the event in which the coin that was picked landed on heads $B$.
We want to find $P$($A$|$B$).
By Bayes' rule, we have that
$$P(A|B)= \frac{P(B|A) \,P(A)}{P(B)}.$$
The event $B$|$A$ is equivalent to the event in which the coin that was picked out of the $6$ landed on heads, given the fact that the fair coin landed on heads. Therefore,
$$P(B|A)=3/6=1/2.$$
Also, $P$($A$) (the probability that the fair coin landed on heads) = $1$/$2$.
Now, we will calculate $P$($B$) using the Law of Total Probability.
$$P(B)=P(A)\,P(B|A)+P(A^C)\,P(B|A^C)$$ $$=1/2*3/6 + 1/2*2/6$$ $$=5/12.$$
Now, we have all the unknowns necessary to solve for
$$P(A|B)=\frac{P(B|A)\,P(A)}{P(B)}=\frac{\frac 12 \frac 12}{\frac {5} {12}}=\frac 35$$