Question

Assume that when adults with smartphones are randomly​ selected, 46​% use them in meetings or classes. If 9 adult smartphone users are randomly​ selected, find the probability that at least 6 of them use their smartphones in meetings or classes.

Answer 1

Answer:

0.18173219.

Step-by-step explanation:

We have been asked to find what will be the probability that at least 6 of 9 adults use their smartphones in meetings or classes.

We will find our answer using Bernoulli's trails.

$_{r}^{n}\textrm{c}\cdot p^{r}\cdot (1-p)^{n-r}$

First of all we will find the probabilities when r is 6, 7,8 and 9 then we will add them all.

When r=6,

$_{6}^{9}\textrm{c}\cdot 0.46^{6}\cdot (1-0.46)^{9-6}$

$\frac{9!}{6!3!} *0.46^{6} *0.54^{3}$

$\frac{9*8*7*6!}{6!*3*2*1} *0.009474296896*0.157464$

$84*0.009474296896*0.157464=0.12532$

Similarly we will find Probabilities when r=7, 8 and 9.

When r=7

$_{7}^{9}\textrm{c}\cdot 0.46^{7}\cdot (1-0.46)^{9-7}$

$\frac{9!}{7!2!} *0.46^{7} *0.54^{2}$

$\frac{9*8*7!}{7!*2*1} *0.00435817657216*0.2916$

$36*0.00435817657216*0.2916=0.04575039$

When r=8

$_{8}^{9}\textrm{c}\cdot 0.46^{8}\cdot (1-0.46)^{9-8}$

$\frac{9!}{8!1!} *0.46^{8} *0.54$

$\frac{9*8!}{8!*1!} *0.00200476*0.54$

$9 *0.00200476*0.54=0.0097431336$

When r=9,

$_{9}^{9}\textrm{c}\cdot 0.46^{9}\cdot (1-0.46)^{9-9}$

$\frac{9!}{9!0!} *0.46^{9} *1$

$1 *0.00092219*1=0.00092219$

Now let us add all the probabilities to get the final answer.

$0.12532+0.04575+0.00974+0.00092219=0.18173219$

Therefore, probability that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219.

Answer 2

The probability that at least $6$ out of $9$ adult use their smartphone in meeting or classes is $\fbox{\begin\\\ 0.1817\\\end{minispace}}$.

Further explanation:

It is given that $46\%$ smartphone user use their smartphone in meetings or classes and at least $6$ out of $9$ adult smartphone user are selected randomly.

Here we will use the concept of Binomial probability.

If an experiment is performed $n$ times and it has only two outcomes that is, “success” and “failure”.So, the probability associated with this experiment is called Binomial probability.

The probability to get exactly $r$ successes in $n$ trial is given as follows,

$\boxed{P=^{n}C_{r}p^{r}(1-p)^{n-r}}$ ......(1)

Here, $P$ is the probability of $r$ successes in $n$ trials.

About $46\%$ smartphone user use their smartphone in meetings or classes that means the probability of the smartphone user use their smartphone in meetings or classes is as follows:

$\begin{aligned}46\%&=\dfrac{46}{100}\\&=0.46\end{aligned}$

The statement, “at least six smartphone user” means six or more smartphone user. We will calculate the probability of the $6,7,8$ and $9$ adult smartphone user.

Substitute $0.46$ for $p$, $9$ for $n$ and $6,7,8,9$ for $r$ in equation (1) to obtain the probability of summation as follows,

$P=^{9}C_{6}(0.46)^{6}(1-0.46)^{9-6}+^{9}C_{7}(0.46)^{7}(1-0.46)^{9-7}+^{9}C_{8}(0.46)^{8}(1-0.46)^{9-8}+\qquad^{9}C_{9}(0.46)^{9}(1-0.46)^{9-9}\\\\P=\dfrac{9!}{6!\cdot(9!-6!)}\cdot(0.46)^{6}\cdot(0.54)^{3}+\dfrac{9!}{7!\cdot(9!-7!)}\cdot(0.46)^{7}\cdot(0.54)^{2}+\dfrac{9!}{8!\cdot(9!-8!)}\cdot(0.46)^{8}\cdot(0.54)^{1}+\dfrac{9!}{9!\cdot(9!-9!)}\cdot(0.46)^{9}\cdot(0.54)^{0}$  

Further solving the above equation as follows,  

$\begin{aligned}P&=84\cdot(0.46)^{6}\cdot(0.54)^{4}+36\cdot(0.46)^{7}\cdot(0.54)^{2}+9\cdot(0.46)^{8}\cdot(0.54)^{1}+1\cdot(0.46)^{9}\cdot(0.54)^{0}\\&=0.1253+0.04575+0.009743+0.00092219\\&=0.1817\end{aligned}$  

Therefore, the probability that at least $6$ out of $9$ adult use their smartphone in meeting or classes is $\boxed{0.1817}$.

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Answer details:

Grade: College

Subject: Mathematics

Chapter: Probability

Keywords:  Probability, numbers, chances, smartphone, meeting, classes, 0.1817, summation, user, equation, selected, Binomial probability, randomly.