A fair coin is flipped 3 times. Probability of all $3$ heads given at least $2$ were heads

Question

Here's a problem from Prof. Blitzstein's Stat 110 textbook. Please see my solution below.

A fair coin is flipped 3 times. The toss results are recorded on separate slips of paper (writing “H” if Heads and “T” if Tails), and the 3 slips of paper are thrown into a hat.

(a) Find the probability that all 3 tosses landed Heads, given that at least 2 were Heads.

(b) Two of the slips of paper are randomly drawn from the hat, and both show the letter H. Given this information, what is the probability that all 3 tosses landed Heads?

The binomial distribution of the number of heads (#H) is as follows:

$\mathbb{P}(\#H=0) = 1/8 $
$\mathbb{P}(\#H=1) = 3/8 $
$\mathbb{P}(\#H=2) = 3/8 $
$\mathbb{P}(\#H=3) = 1/8 $

a) Probability of 3 heads given at least 2 heads = $\mathbb{P}(\#H=3)/\mathbb{P}(\#H \ge 2 ) = (1/8)/(1/2)=1/4$

b) But I'm not able to understand how (b) is different from (a). To me it's the same problem and has the same answer.

Can anyone help please?

Answer 1

Maybe it is the same problem. But intuition can be unreliable in these matters, so let us compute.

Let $A$ be the event "three heads" and let $B$ be the event "first two slips drawn were heads." We want $\Pr(A\mid B)$, which is $\Pr(A\cap B)/\Pr(B)$. Now let us compute $\Pr(B)$.

The event $B$ can happen in two ways: (i) all three were heads and (ii) two were heads, and the first two slips drawn were heads. The probability of (1) is $1/8$.

For (ii), the probability of $2$ heads is $3/8$. Given we have a total of $2$ heads, the probability the first two slips are head is $(2/3)(1/2)$. So the probability of (ii) $(3/8)(2/3)(1/2)$, that is, $1/8$.

It follows that $\Pr(A\mid B)=\frac{1/8}{1/8+1/8}=\frac{1}{2}$.

Remark: In retrospect, this should be almost obvious, for putting aside and then looking is probabilistically equivalent to looking. So here we are basically calculating the probability the third toss is head given the first two are. This is $1/2$.

Answer 2

(b) involves two processes: First flip the coins, then read two of the results.

So, for instance, if you flip exactly two coins, what is the (conditional )probability that you the two papers read are those of the two heads?

Let $H_f$ be the count of heads flipped and $H_r$ the count of heads read.   Use Bayes' Rule and Law of Total Probability.

$$\begin{align}\mathsf P(H_f = 3\mid H_r=2) =&~ \dfrac{\mathsf P(H_r=2\mid H_f=3)~\mathsf P(H_f=3)}{\sum_{x=0}^3\mathsf P(H_r=2\mid H_f=x)~\mathsf P(H_f=x)} \end{align}$$

(Note: several of the terms of the sum will be zero because it is impossible to read 2 if you have thrown less than that.)

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PS: As Andre mentions in his answer, because each paper belongs to a specific coin, this is practically the same as flipping three coins, noting that the first two looked at are heads, and asking for the probability that the third is also heads.

Answer 3

Perhaps it's worth emphasising that the difference is that in (a) the condition is that any $2$ coins landed heads and in (b) the condition is that $2$ particular coins landed heads. Condition (b) is more specific; it leaves only $2$ possible configurations, whereas (a) leaves $4$. Since in each case you're looking for the probability of one of these configurations occurring, the probabilities are $\frac12$ and $\frac14$.