Question

a coin is tossed three times. an outcome is represented by a string of the sort htt (meaning a head on the first toss, followed by two tails). the 8 outcomes are listed in the table below. note that each outcome has the same probability. for each of the three events in the table, check the outcome(s) that are contained in the event. then, in the last column, enter the probability of the event. outcomes probability hht hth tht tth thh htt ttt hhh event a: a tail on the first toss or the third toss (or both) event b: two or more tails event c: a head on each of the first two tosses

Explanation:

Step1: Analyze Event A

An outcome has a tail on the first toss (first character 'T') or tail on the third toss (third character 'T'). Let's list the outcomes:

• THT (T first, T third)
• TTH (T first, T second)
• THH (T first, H third)
• HTT (H first, T third)
• TTT (T first, T third)

So that's 5 outcomes? Wait, wait: Wait, first toss T: THT, TTH, THH, TTT. Third toss T: HTT, THT, TTH, TTT. Wait, using inclusion - exclusion: number of outcomes with T first: 4 (THT, TTH, THH, TTT), number with T third: 4 (HTT, THT, TTH, TTT), number with both: 3 (THT, TTH, TTT). So total is 4 + 4 - 3 = 5? Wait no, let's list all 8 outcomes:

1. HHT: first H, third T? No, first H, third T? Wait HHT: first H, second H, third T. So third toss T. Wait, I made a mistake earlier. Let's list each outcome:

• HHT: first H, third T → included (third T)
• HTH: first H, third H → not included (neither first T nor third T)
• THT: first T, third T → included (both)
• TTH: first T, third T → included (both)
• THH: first T, third H → included (first T)
• HTT: first H, third T → included (third T)
• TTT: first T, third T → included (both)
• HHH: first H, third H → not included

So let's check each:

• HHT: third T → yes
• HTH: no
• THT: yes (first T or third T)
• TTH: yes
• THH: yes (first T)
• HTT: yes (third T)
• TTT: yes
• HHH: no

So the included outcomes are HHT, THT, TTH, THH, HTT, TTT. Wait that's 6? Wait HHT: first H, third T → yes (third T). THT: yes. TTH: yes. THH: yes. HTT: yes. TTT: yes. And what about HTH? No. HHH? No. So that's 6 outcomes. Wait, let's count again:
Outcomes:

1. HHT: first H, third T → included (third T)
2. HTH: first H, third H → not included
3. THT: first T, third T → included
4. TTH: first T, third T → included
5. THH: first T, third H → included (first T)
6. HTT: first H, third T → included (third T)
7. TTT: first T, third T → included
8. HHH: first H, third H → not included

So that's 6 outcomes. So probability is 6/8 = 3/4. Wait, maybe my initial inclusion - exclusion was wrong. Let's do it properly:
Number of outcomes with first toss T: THT, TTH, THH, TTT → 4
Number with third toss T: HHT, THT, TTH, HTT, TTT → 5? Wait no, third toss T: HHT (third T), THT (third T), TTH (third T), HTT (third T), TTT (third T) → 5. First toss T: THT, TTH, THH, TTT → 4. Both: THT, TTH, TTT → 3. So by inclusion - exclusion, 4+5 - 3=6. Yes, that's correct. So Event A has 6 outcomes. Since each outcome has probability 1/8, probability of A is 6/8 = 3/4.

Step2: Analyze Event B

Event B: Two or more tails. Let's count the number of tails in each outcome:

• HHT: 1 tail (third)
• HTH: 1 tail (second)
• THT: 2 tails (first and third)
• TTH: 2 tails (first and second)
• THH: 1 tail (first)
• HTT: 2 tails (second and third)
• TTT: 3 tails
• HHH: 0 tails

So outcomes with two or more tails: THT (2), TTH (2), HTT (2), TTT (3). So that's 4 outcomes. So probability is 4/8 = 1/2.

Step3: Analyze Event C

Event C: A head on each of the first two tosses. So first toss H and second toss H. Let's check each outcome:

• HHT: first H, second H → yes
• HTH: first H, second T → no
• THT: first T, second H → no
• TTH: first T, second T → no
• THH: first T, second H → no
• HTT: first H, second T → no
• TTT: first T, second T → no
• HHH: first H, second H → yes

So the outcomes are HHT and HHH. So 2 outcomes. Probability is 2/8 = 1/4.

Answer:

Event A: Probability $\frac{6}{8}=\frac{3}{4}$; Event B: Probability $\frac{4}{8}=\frac{1}{2}$; Event C: Probability $\frac{2}{8}=\frac{1}{4}$

(Note: For the checkboxes, for Event A: check HHT, THT, TTH, THH, HTT, TTT; for Event B: check THT, TTH, HTT, TTT; for Event C: check HHT, HHH)