Question

lecture 13 quiz 2
above, you are given the outcomes of a laplace probability model for flipping a fair coin three times. if the last two coin flips are the same, what is closest to the probability that you get heads (h) exactly twice?
▶ a. 0
▶ b. 1/6
▶ c. 2/6
▶ d. 3/6
▶ e. 4/6
▶ f. 1
▶ g. not enough information given

Explanation:

Step1: Identify relevant outcomes

We first filter all 3-flip outcomes where the last two flips are the same. The valid outcomes are: HHH, HTT, THH, TTT, HHT, TTH (wait, correction: from the given set, the outcomes with last two same are HHH, HTT, THH, TTT, HHT, TTH? No, let's list the 8 total outcomes and select those where 2nd and 3rd are equal:
- HHH (last two H,H: same)
- HHT (last two H,T: no)
- HTH (last two T,H: no)
- HTT (last two T,T: same)
- THH (last two H,H: same)
- THT (last two H,T: no)
- TTH (last two T,H: no)
- TTT (last two T,T: same)
Wait, the given box has: HHH, HTH, THH, TTH, HHT, HTT, THT, TTT. So the outcomes where last two flips are same are: $\text{HHH, HTT, THH, TTT, HHT? No, HHT last two are H,T: no. Correct valid set: } S = \{\text{HHH, HTT, THH, TTT}\}$? No, wait: last two flips same means flip2=flip3.
- HHH: flip2=H, flip3=H: yes
- HTH: flip2=T, flip3=H: no
- THH: flip2=H, flip3=H: yes
- TTH: flip2=T, flip3=H: no
- HHT: flip2=H, flip3=T: no
- HTT: flip2=T, flip3=T: yes
- THT: flip2=H, flip3=T: no
- TTT: flip2=T, flip3=T: yes
So total valid outcomes: $n(S) = 4$ (HHH, HTT, THH, TTT)

Step2: Identify favorable outcomes

Now find outcomes in $S$ with exactly 2 Heads.
- HHH: 3 Heads: no
- HTT: 1 Head: no
- THH: 2 Heads: yes
- TTT: 0 Heads: no
Wait, correction: wait THH has two Heads, is there another? No, let's recheck: the only outcome in the valid set with exactly 2 H is THH? Wait no, wait HHT is not in the valid set (last two flips H and T, not same). Wait no, wait the condition is "last two coin flips are the same"—so flip 2 and 3 are same. An outcome with exactly 2 H would need two H and one T, with flip2=flip3.
If flip2=flip3=H: then flip1 must be T to get exactly 2 H (THH)
If flip2=flip3=T: then flip1 would need to be H to get 1 H (HTT), which is not 2 H.
So favorable outcomes: $n(F) = 1$? No, wait wait HHH has 3 H, THH has 2 H. Wait wait, no: wait "exactly two Heads"—so count of H is 2. THH has two H, is there another? Let's see: is there an outcome with flip2=flip3, and exactly two H?
Suppose flip1=H, flip2=H, flip3=H: 3 H (not exactly 2)
flip1=T, flip2=H, flip3=H: 2 H (yes)
flip1=H, flip2=T, flip3=T: 1 H (no)
flip1=T, flip2=T, flip3=T: 0 H (no)
Wait, but wait the problem says "the outcomes of a Laplace probability model for flipping a fair coin three times"—so all 8 outcomes are equally likely, and we are conditioning on last two flips same.
Wait, wait I made a mistake: HHT: last two flips are H and T, not same, so excluded. HTH: last two T and H, excluded. THT: last two H and T, excluded. TTH: last two T and H, excluded. So valid conditional space is 4 outcomes, as above.

Step3: Calculate conditional probability

Conditional probability formula: $P(F|S) = \frac{n(F)}{n(S)}$
Substitute values: $\frac{1}{4}$? No, wait wait—wait THH is one, is there another? Wait no: wait what about HHT? No, HHT last two flips are H and T, which are not same, so it's not in the conditional set. Wait wait, no: "last two coin flips are the same"—flip 2 and flip 3, right? Yes, "last two" means the second and third flip in three flips.
Wait wait, wait another way: list all 8 outcomes, mark those with last two same:
1. HHH: yes (last two H,H)
2. HHT: no (last two H,T)
3. HTH: no (last two T,H)
4. HTT: yes (last two T,T)
5. THH: yes (last two H,H)
6. THT: no (last two H,T)
7. TTH: no (last two T,H)
8. TTT: yes (last two T,T)
So 4 outcomes in conditional space. Now how many have exactly 2 H? Only THH (T, H, H) has exactly two H. HHH has three H, which is not exactly two. HTT has one H, TTT has zero. So $n(F)=1$, $n(S)=4$, so probability is $\frac{1}{4}=0.25$. Wait but the options are 0, 1/8, 2/8, 3/8, 4/8, 1, not enough info. Wait 1/4 is equal to 2/8. Oh! $\frac{1}{4} = \frac{2}{8}$.
Wait wait, did I miscount favorable outcomes? Wait is there another outcome with exactly two H and last two flips same? Let's see: to have exactly two H, and last two same. If last two are H, then first flip is T (THH: two H). If last two are T, then first flip would need to be H and T? No, if last two are T, then total H would be 1 (HTT), which is not two. Wait wait—wait, is "last two coin flips are the same" referring to flip 1 and 2? No, "last two" of three flips are flip 2 and 3. Wait no, wait "last two" could be flip 1 and 2? No, "last two" in three flips are the second and third. Wait wait, maybe I misinterpreted "last two coin flips are the same" as the two final flips being identical, but maybe it means that the result of the last flip is the same as the one before it (i.e., flip2=flip3), which is what I did.
Wait wait another approach: use conditional probability formula.
Let A = event that last two flips are same, B=event that exactly two H.
$P(B|A) = \frac{P(A \cap B)}{P(A)}$
$P(A)$: probability last two flips same. For fair coin, flip2 and flip3 are same with probability $\frac{1}{2}$ (HH or TT, each with probability $\frac{1}{4}$, sum to $\frac{1}{2}$)
$P(A \cap B)$: probability last two flips same AND exactly two H. This happens only when flip2=flip3=H, and flip1=T: probability $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$? No, wait: flip1=T, flip2=H, flip3=H: probability $\frac{1}{8}$. Is there another case? If flip2=flip3=T, then to have exactly two H, we would need two H, but flip2 and 3 are T, so flip1 would need to be two H, which is impossible. So $P(A \cap B) = \frac{1}{8}$
Then $P(B|A) = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} = \frac{2}{8}$
Ah, yes, $\frac{1}{4}$ is equal to $\frac{2}{8}$, which is option C.

Step4: Match to option

$\frac{1}{4} = \frac{2}{8}$, which is option C.

Answer:

C. 2/8