Question

find the binomial distribution for flipping a coin 3 times, where \heads\ is a success. p(k successes) = _nc_k p^k(1 - p)^(n - k) p(0 successes) = _3c_0(0.5)^0(0.5)^3 = 0.125 p(1 success) = 0.375 p(2 successes) = 0.375 p(3 successes) = 0.125 complete what is the probability that you will get \heads\ no more than once out of 3 flips? p(0 successes) + p(1 success) =

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(k\text{ successes})={}_nC_kp^k(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial. Here, $n = 3$ (number of coin - flips) and $p=0.5$ (probability of getting heads in a single coin - flip).

Step2: Calculate $P(0\text{ successes})+P(1\text{ success})$

We know that $P(0\text{ successes})={}_3C_0(0.5)^0(0.5)^3=\frac{3!}{0!(3 - 0)!}\times1\times(0.5)^3 = 0.125$ and $P(1\text{ success})={}_3C_1(0.5)^1(0.5)^2=\frac{3!}{1!(3 - 1)!}\times0.5\times(0.5)^2=3\times0.5\times0.25 = 0.375$.

Step3: Sum the probabilities

$P(0\text{ successes})+P(1\text{ success})=0.125 + 0.375=0.5$

Answer:

$0.5$