Question

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

Explanation:

Step1: Recall combination formula

The combination formula is $_{n}C_{k}=\frac{n!}{k!(n - k)!}$, and for a fair - coin $p = 0.5$, $n=3$.

Step2: Calculate $P(0\ successes)$

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

Step3: Calculate $P(1\ success)$

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

Step4: Calculate $P(2\ successes)$

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

Step5: Calculate $P(3\ successes)$

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

Answer:

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