# A box contains three coins; one coin is two-headed, one coin is fair, and another coin is weighted so that the probability of heads appearing is 1/3.

A coin is randomly selected and flipped.

-Two-headed coin (100% chance in getting heads)

-Fair coin (50% chance in getting heads)

-Weighted Coin (33.3333% chance in getting heads)

(a) What is the probability (p) that heads appear?

-Given the info above, I got 33/54.

(b) What is the probability that the coin is two-headed if the coin selected is known to be head?

I don't understand how to solve for the probability using Bayes' Theorem. Can someone please guide me through the logic of solving such problems? Thank you!

• Have you done part "a"? That does not require Bayes' Theorem; just a clear notion of the probability space and the probability of each outcome in that space. – John Hughes Aug 9 '17 at 17:59
• "A coin is randomly selected" have the $3$ distinghuishable coins equal chances to be selected? – drhab Aug 9 '17 at 18:01

## 1 Answer

Let $H$ be the event that heads appears, $T$ the event that the two-headed coin is drawn, $F$ the event that the fair coin is drawn, and $W$ the event that the weighted coin is drawn.

For part (a), we just need to compute $P(H)$. Since $T$, $F$, and $W$ partition the sample space (i.e., they are pairwise disjoint and their union accounts for all possibilities), we can use the law of total probability to get $$P(H) = P(H \mid T)P(T) + P(H\mid F)P(F) + P(H\mid W)P(W) = (1)(1/3) + (1/2)(1/3) + (1/3)(1/3) = 33/54.$$

Now, for part (b), we use Bayes' theorem to compute $$P(T \mid H) = \frac{P(H\mid T)P(T)}{P(H)} = \frac{(1)(1/3)}{(33/54)}=6/11.$$

To better understand what is going on with Bayes' theorem, I find it useful to derive the formula above by using the definition of conditional probability: $$P(A\mid B) = \frac{P(A\cap B)}{P(B)}.$$ This definition shows that $P(A\cap B) = P(A\mid B)P(B)$, and so by the symmetry of intersection, $P(A\cap B)=P(B\cap A) = P(B\mid A)P(A)$. This yields the formula $$P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B\mid A)P(A)}{P(B)}.$$ Thus, in a way, we sort of "reverse" our assumption from $B$ to $A$.

Interestingly enough, this is also how the law of total probability is derived. If $X_1,\ldots,X_n$ partition the sample space, then we have $$P(A) = P(A\cap X_1) + \cdots + P(A\cap X_n) = P(A\mid X_1)P(X_1)+\cdots +P(A\mid X_n)P(X_n).$$

• Ah, just wondering...I did part b based on the steps that you showed, but up to the last step I got (1)(1/3)(33/54)=6/11. Did I make a mistake or did you? – Sarah Henner Aug 9 '17 at 18:20
• @SarahHenner My mistake! – John Griffin Aug 9 '17 at 18:26