Question

find the probability that at most 2 heads occured when 5 fair coins are tossed.
the probability that at most 2 are heads is
(type an integer or decimal rounded to the nearest thousandth as needed.)

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 5$ (number of coin - tosses), $p=\frac{1}{2}$ (probability of getting a head in a single coin - toss).

Step2: Calculate $P(X = 0)$

$C(5,0)=\frac{5!}{0!(5 - 0)!}=\frac{5!}{5!}=1$, $P(X = 0)=C(5,0)\times(\frac{1}{2})^{0}\times(1-\frac{1}{2})^{5 - 0}=1\times1\times(\frac{1}{2})^{5}=\frac{1}{32}$.

Step3: Calculate $P(X = 1)$

$C(5,1)=\frac{5!}{1!(5 - 1)!}=\frac{5!}{1!4!}=5$, $P(X = 1)=C(5,1)\times(\frac{1}{2})^{1}\times(1-\frac{1}{2})^{5 - 1}=5\times\frac{1}{2}\times(\frac{1}{2})^{4}=\frac{5}{32}$.

Step4: Calculate $P(X = 2)$

$C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10$, $P(X = 2)=C(5,2)\times(\frac{1}{2})^{2}\times(1-\frac{1}{2})^{5 - 2}=10\times\frac{1}{4}\times(\frac{1}{2})^{3}=\frac{10}{32}$.

Step5: Calculate $P(X\leq2)$

$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)=\frac{1 + 5+10}{32}=\frac{16}{32}=0.500$.

Answer:

$0.500$