Probability Question: 6 coins, 3 two tails, 2 two heads, 1 fair coin

Question

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!

Answer 1

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

$$P(A|B)=\frac{P(B|A)\,P(A)}{P(B)}=\frac{\frac 12 \frac 12}{\frac {5} {12}}=\frac 35$$

Answer 2

The situation can be modeled as follows. We fix numbers for the coins. These are:

• coin 1: TT
• coin 2: TT
• coin 3: TT
• coin 4: HH
• coin 5: HH
• coin 6: HT

We flip them, and see

• either TTTHHH,
• or TTTHHT.

This is the first step, where some randomness occurs. So we draw a tree to suggest this situation, and we also draw the probabilities for the one or the other choice, both $1/2$:

    TTTHHH
   /
  / 1/2
 /
o
 \
  \ 1/2
   \
    TTTHHT

Now there is a fair choice (extracted humanly from the text) to take one of the coins, show it, and hide the others. Let us display the six choices for each branch, each comes further with probability $1/6$.

            -- [T]TTHHH
           /-- T[T]THHH
          /--- TT[T]HHH
    TTTHHH---- TTT[H]HH
   /      \--- TTTH[H]H
  / 1/2    \-- TTTHH[H]
 /
o
 \          -- [T]TTHHT
  \ 1/2    /-- T[T]THHT
   \      /--- TT[T]HHT
    TTTHHT---- TTT[H]HT
          \--- TTTH[H]T
           \-- TTTHH[T]

Each of the $12$ cases is equally probable.

Above, the one visible coin is [X], with X being either T or H. We get the information that the one shown coin is [H]. To get the conditional probability for the last coin to be H, conditioned by the even that we see an [H], we have to restrict to the corresponding cases... The tree is "cleaned" to offer only:

            -- 
           /-- 
          /--- 
    TTTHHH---- TTT[H]HH
   /      \--- TTTH[H]H
  /        \-- TTTHH[H]
 /
o
 \          -- 
  \        /-- 
   \      /--- 
    TTTHHT---- TTT[H]HT
          \--- TTTH[H]T
           \-- 

Each of the remained $5$ cases is equally probable. And the conditional probability to see an H on the last place, the place of the fair coin, is $\bbox[yellow]{3/5}$.

Answer 3

We don’t know if someone picked which coins to remove or if it was done randomly. If I always remove the fair coin then the probability is 0.5. If I never remove the fair coin then the probability is 1. If I always keep a random head (one of two or three) then the probability is 0.5.

If I keep one coin at random: The probability for head is 2/6 if the fair coin is tail, and 3/6 if the fair coin is head. So 3 out of 5 cases the fair coin was heads.

Answer 4

Note has to be taken of the wording.
We aren't asked the probability that the head seen is from the fair coin, rather "What is the probability that the fair coin has landed on heads?"

We can reduce the sample space to coins with "heads possible"

[HH] [HH] [HT] has $5$ equiprobable ways heads could be seen

In $4$ of them, P(the fair coin shows heads ) $= \Large\frac12$
and in the last case, it is the fair coin that we saw

Thus $Pr = \Large\frac45\cdot\frac12 + \Large\frac15\cdot1= \Large\frac35$