Question

according to a recent survey, 80% of high school students have their own cell phone. suppose you select 10 high school students at random. determine each probability. round your answers to the nearest tenth of a percent if necessary. sample problem p(8 of the students have cell phones) =$_{10}c_{8}cdot(\frac{8}{10})^{8}(\frac{2}{10})^{2}$≈45(0.0067)≈0.3015≈30.2% p(3 of the students do not have cell phones) > enter the answer in the space provided. use numbers instead of words.

Explanation:

Step1: Identify the 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 $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$ for the first part

For the probability that 8 of the 10 students have cell - phones, $n = 10$, $k = 8$, and $p=0.8$. First, calculate the combination $C(10,8)=\frac{10!}{8!(10 - 8)!}=\frac{10!}{8!2!}=\frac{10\times9}{2\times1}=45$. Then, $p^{k}=(0.8)^{8}$ and $(1 - p)^{n - k}=(0.2)^{2}$. So, $P(X = 8)=C(10,8)\times(0.8)^{8}\times(0.2)^{2}=45\times(0.8)^{8}\times(0.2)^{2}\approx45\times0.16777216\times0.04\approx0.30199\approx30.2\%$.

Step3: Determine the values of $n$, $k$, and $p$ for the second part

For the probability that 3 of the 10 students do not have cell - phones, the probability that a student does not have a cell - phone is $1 - p = 0.2$. Here, $n = 10$, $k = 3$, and $p = 0.2$. Calculate the combination $C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120$. Then, $p^{k}=(0.2)^{3}$ and $(1 - p)^{n - k}=(0.8)^{7}$. So, $P(X = 3)=C(10,3)\times(0.2)^{3}\times(0.8)^{7}=120\times0.008\times0.2097152\approx0.20133\approx20.1\%$.

Answer:

20.1%