Question

find the binomial distribution for flipping a coin 3 times, where \heads\ is a success.
\\(p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}\\)
\\(p(0\text{ successes}) = _3c_0(0.5)^0(0.5)^3=)
\\(p(1\text{ success}) =)
\\(p(2\text{ successes}) =)
\\(p(3\text{ successes}) =
done

Explanation:

Step1: Recall combination formula

The combination formula is $_{n}C_{k}=\frac{n!}{k!(n - k)!}$. Here $n = 3$ (number of coin - flips) and for different cases $k$ varies. Also, $p=0.5$ (probability of getting heads in a single coin - flip) and $1 - p = 0.5$.

Step2: Calculate $P(0\text{ successes})$

$_{3}C_{0}=\frac{3!}{0!(3 - 0)!}=\frac{3!}{3!}=1$, $p = 0.5$, $n = 3$, $k = 0$. So $P(0\text{ successes})=_{3}C_{0}(0.5)^{0}(0.5)^{3}=1\times1\times0.125 = 0.125$.

Step3: Calculate $P(1\text{ success})$

$_{3}C_{1}=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=\frac{3\times2!}{2!}=3$. Then $P(1\text{ success})=_{3}C_{1}(0.5)^{1}(0.5)^{2}=3\times0.5\times0.25 = 0.375$.

Step4: Calculate $P(2\text{ successes})$

$_{3}C_{2}=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!}=3$. So $P(2\text{ successes})=_{3}C_{2}(0.5)^{2}(0.5)^{1}=3\times0.25\times0.5 = 0.375$.

Step5: Calculate $P(3\text{ successes})$

$_{3}C_{3}=\frac{3!}{3!(3 - 3)!}=\frac{3!}{3!0!}=1$. Then $P(3\text{ successes})=_{3}C_{3}(0.5)^{3}(0.5)^{0}=1\times0.125\times1 = 0.125$.

Answer:

$P(0\text{ successes}) = 0.125$
$P(1\text{ success}) = 0.375$
$P(2\text{ successes}) = 0.375$
$P(3\text{ successes}) = 0.125$