Answer: 0.18173219
The probability is that at least 6 of 9 adults use their smartphones in meetings or classes is 0.18173219
Hope this helps!!! Good luck!!! ;)
Using the binomial distribution, it is found that there is a 0.8999 = 89.99% probability that at least 3 of them use their smartphones in meetings or classes.
For each adult, there are only two possible outcomes. Either they use their smartphones in meetings or classes, or they do not. The probability of an adult using their smartphones in meetings or classes is independent of any other adult, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this problem:
The desired probability is:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.49)^{0}.(0.51)^{9} = 0.0023[/tex]
[tex]P(X = 1) = C_{9,1}.(0.49)^{1}.(0.51)^{8} = 0.0202[/tex]
[tex]P(X = 2) = C_{9,2}.(0.49)^{2}.(0.51)^{7} = 0.0776[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0023 + 0.0202 + 0.0776 = 0.1001[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.1001 = 0.8999[/tex]
0.8999 = 89.99% probability that at least 3 of them use their smartphones in meetings or classes.
A similar problem is given at https://brainly.com/question/24863377
