Question

thth
thtt
tthh
ttht
ttth
tttt
what is the probability of getting the same number of heads and tails?
\\(\frac{1}{8}\\)
\\(\frac{1}{4}\\)
\\(\frac{3}{8}\\)
\\(\frac{3}{4}\\)

Explanation:

Step1: Count total outcomes

Total listed outcomes: $6$
Wait, correction: For 4 coin flips, total possible outcomes are $2^4=16$, but the given list is incomplete. However, we need outcomes with 2 heads (same number of heads and tails, since 4 flips mean 2 H, 2 T).

Step2: Identify valid outcomes

From the list, outcomes with 2 H and 2 T: THHT, TTHH → 2 outcomes.
Wait, correction: Wait, full set of 4-flip outcomes with 2H: $\binom{4}{2}=6$. But the given list only shows 6 outcomes, so we use the given sample.

Step3: Calculate probability

Probability = $\frac{\text{Valid outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3}$? No, wait, no—wait, the list is partial. Wait, no, the question uses the provided list as the sample space. Let's recheck:
Listed outcomes:

1. THTH (2H, 2T)
2. THTT (1H, 3T)
3. TTHH (2H, 2T)
4. TTHT (1H, 3T)
5. TTTH (1H, 3T)
6. TTTT (0H, 4T)

Step1: Count total sample size

Total outcomes: $6$

Step2: Count favorable outcomes

Favorable (2H,2T): THTH, TTHH → $2$

Step3: Compute probability

$\text{Probability} = \frac{2}{6} = \frac{1}{3}$? No, this is not an option. Wait, I made a mistake: 4 coin flips, same number of heads and tails means 2 heads and 2 tails. The full sample space has $2^4=16$ outcomes. The number of 2H outcomes is $\binom{4}{2}=6$. So probability is $\frac{6}{16}=\frac{3}{8}$. The given list is a partial list, but the question is about the theoretical probability, not the listed sample.

Step1: Define total possible outcomes

For 4 coin flips: $2^4 = 16$

Step2: Count favorable outcomes

Number of ways to get 2 heads: $\binom{4}{2} = \frac{4!}{2!2!} = 6$

Step3: Calculate probability

$\text{Probability} = \frac{6}{16} = \frac{3}{8}$

Answer:

$\boldsymbol{\frac{3}{8}}$ (Option: $\boldsymbol{\frac{3}{8}}$)