Question

1. the possible outcomes for flipping three coins are (hhh, hht, hth, htt, thh, tht, tth, ttt). what is the probability of the coin landing on at least two heads when it is flipped? a $\frac{1}{2}$ b $\frac{3}{8}$ c $\frac{7}{8}$ d $\frac{1}{8}$

Explanation:

Step1: Identify total outcomes

When flipping three coins, the total number of possible outcomes is \(2^3 = 8\) (since each coin has 2 outcomes: heads (H) or tails (T)). The given outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.

Step2: Identify favorable outcomes

We need the outcomes with at least two heads. "At least two heads" means two heads or three heads.

• Outcomes with two heads: HHT, HTH, THH (3 outcomes)
• Outcomes with three heads: HHH (1 outcome)

So total favorable outcomes \(= 3 + 1 = 4\)? Wait, no, wait: Wait, HHT, HTH, THH are two heads, HHH is three heads. Wait, 3 + 1 = 4? Wait, no, let's list again:
Wait the possible outcomes for three coins:

1. HHH (3 heads)
2. HHT (2 heads)
3. HTH (2 heads)
4. HTT (1 head)
5. THH (2 heads)
6. THT (1 head)
7. TTH (1 head)
8. TTT (0 heads)

So outcomes with at least two heads (2 or 3 heads) are HHH, HHT, HTH, THH. That's 4 outcomes? Wait but the options have 3/8, 1/2, etc. Wait, wait, maybe I made a mistake. Wait 2 heads: HHT, HTH, THH (3 outcomes), 3 heads: HHH (1 outcome). So total 4? But 4/8 = 1/2? But option B is 3/8, A is 1/2. Wait, maybe the problem is "at least two heads" – wait, maybe the original problem's possible outcomes are listed as HHH, HHT, HTH, HTT, THH, THT, TTH, TTT (8 outcomes). Wait, let's check again:

Wait, when flipping three coins, the number of ways to get 2 heads: \(\binom{3}{2}=\frac{3!}{2!(3 - 2)!}=3\), and 3 heads: \(\binom{3}{3}=1\). So total favorable is \(3 + 1 = 4\). So probability is \(\frac{4}{8}=\frac{1}{2}\)? But wait, the options: A is 1/2, B is 3/8, C is 7/8, D is 1/8. Wait, maybe I misread the problem. Wait the problem says "at least two heads" – maybe the user made a typo, or maybe I misread the outcomes. Wait the given outcomes are (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT) – 8 outcomes. So at least two heads: HHH (3), HHT (2), HTH (2), THH (2) – that's 4 outcomes. So 4/8 = 1/2, which is option A? Wait but maybe the problem is "at least one head"? No, the question is "at least two heads". Wait, maybe the original problem's possible outcomes are different? Wait, no, three coins: 8 outcomes. Wait, let's recalculate:

Number of ways to get 2 heads: \(\binom{3}{2}=3\) (HHT, HTH, THH)
Number of ways to get 3 heads: \(\binom{3}{3}=1\) (HHH)
Total favorable: \(3 + 1 = 4\)
Total outcomes: 8
Probability: \(\frac{4}{8}=\frac{1}{2}\)

Wait, but the option B is 3/8. Wait, maybe the problem is "at least one head"? No, "at least two". Wait, maybe I made a mistake in counting. Wait HHH: 3 heads, HHT: 2, HTH: 2, THH: 2, HTT:1, THT:1, TTH:1, TTT:0. So at least two heads: HHH, HHT, HTH, THH – 4 outcomes. 4/8 = 1/2, which is option A.

Wait, but maybe the problem was "at least one head"? No, the question is "at least two heads". Alternatively, maybe the possible outcomes are listed as HHH, HHT, HTH, HTT, THH, THT, TTH (7 outcomes)? No, the user wrote "The possible outcomes for flipping three coins are (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)" – 8 outcomes. So the calculation is 4/8 = 1/2.

Answer:

A. \(\frac{1}{2}\)