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: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.[tex]_{r}^{n}\textrm{c}\cdot p^{r}\cdot (1-p)^{n-r}[/tex]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,[tex]_{6}^{9}\textrm{c}\cdot 0.46^{6}\cdot (1-0.46)^{9-6}[/tex][tex]\frac{9!}{6!3!} *0.46^{6} *0.54^{3}[/tex][tex]\frac{9*8*7*6!}{6!*3*2*1} *0.009474296896*0.157464[/tex][tex]84*0.009474296896*0.157464=0.12532[/tex]Similarly we will find Probabilities when r=7, 8 and 9.When r=7[tex]_{7}^{9}\textrm{c}\cdot 0.46^{7}\cdot (1-0.46)^{9-7}[/tex][tex]\frac{9!}{7!2!} *0.46^{7} *0.54^{2}[/tex][tex]\frac{9*8*7!}{7!*2*1} *0.00435817657216*0.2916[/tex][tex]36*0.00435817657216*0.2916=0.04575039[/tex]When r=8[tex]_{8}^{9}\textrm{c}\cdot 0.46^{8}\cdot (1-0.46)^{9-8}[/tex][tex]\frac{9!}{8!1!} *0.46^{8} *0.54[/tex][tex]\frac{9*8!}{8!*1!} *0.00200476*0.54[/tex][tex]9 *0.00200476*0.54=0.0097431336[/tex]When r=9,[tex]_{9}^{9}\textrm{c}\cdot 0.46^{9}\cdot (1-0.46)^{9-9}[/tex][tex]\frac{9!}{9!0!} *0.46^{9} *1[/tex][tex]1 *0.00092219*1=0.00092219[/tex]Now let us add all the probabilities to get the final answer.[tex]0.12532+0.04575+0.00974+0.00092219=0.18173219[/tex]Therefore, probability that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219.