To solve the problem of finding the binomial distribution for flipping a coin 3 times, where "heads" is considered a success, we will use the binomial probability formula:
[tex]\[
P(k \text{ successes}) = \binom{n}{k} p^k (1-p)^{n-k}
\][/tex]
Here:
- [tex]\(n\)[/tex] is the number of trials (3 coin flips),
- [tex]\(k\)[/tex] is the number of successes (which can be 0, 1, 2, or 3),
- [tex]\(p\)[/tex] is the probability of success on a single trial (for heads in a fair coin, [tex]\(p = 0.5\)[/tex]),
- [tex]\(\binom{n}{k}\)[/tex] (read as "n choose k") is the binomial coefficient.
Let's calculate the probabilities for each value of [tex]\(k\)[/tex].
### 1. [tex]\(P(0 \text{ successes})\)[/tex]
[tex]\[
P(0 \text{ successes}) = \binom{3}{0} (0.5)^0 (0.5)^{3-0}
\][/tex]
[tex]\[
P(0 \text{ successes}) = 1 \cdot 1 \cdot (0.5)^3 = 1 \cdot 1 \cdot 0.125 = 0.125
\][/tex]
### 2. [tex]\(P(1 \text{ success})\)[/tex]
[tex]\[
P(1 \text{ success}) = \binom{3}{1} (0.5)^1 (0.5)^{3-1}
\][/tex]
[tex]\[
P(1 \text{ success}) = 3 \cdot 0.5 \cdot (0.5)^2 = 3 \cdot 0.5 \cdot 0.25 = 0.375
\][/tex]
### 3. [tex]\(P(2 \text{ successes})\)[/tex]
[tex]\[
P(2 \text{ successes}) = \binom{3}{2} (0.5)^2 (0.5)^{3-2}
\][/tex]
[tex]\[
P(2 \text{ successes}) = 3 \cdot (0.5)^2 \cdot 0.5 = 3 \cdot 0.25 \cdot 0.5 = 0.375
\][/tex]
### 4. [tex]\(P(3 \text{ successes})\)[/tex]
[tex]\[
P(3 \text{ successes}) = \binom{3}{3} (0.5)^3 (0.5)^{3-3}
\][/tex]
[tex]\[
P(3 \text{ successes}) = 1 \cdot (0.5)^3 \cdot 1 = 1 \cdot 0.125 \cdot 1 = 0.125
\][/tex]
Thus, the binomial distribution for flipping a coin 3 times, considering heads as a success, is as follows:
[tex]\[
P(0 \text{ successes}) = 0.125
\][/tex]
[tex]\[
P(1 \text{ success}) = 0.375
\][/tex]
[tex]\[
P(2 \text{ successes}) = 0.375
\][/tex]
[tex]\[
P(3 \text{ successes}) = 0.125
\][/tex]
