# Assume that when adults with smartphones are randomly​ selected, 55​%

Assume that when adults with smartphones are randomly​ selected, 55​% use them in meetings or classes. If 11 adult smartphone users are randomly​ selected, find the probability that fewer than 4 of them use their smartphones in meetings or classes.
The probability is nothing.

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stomachdm

Step 1
Given Information:
55% of adults use their smartphones in meetings or classes. i.e., $$\displaystyle{p}={0.55}$$
If $$\displaystyle{n}={11}$$ adult smartphone users are randomly selected, to find the probability that fewer than 4 of them use their smartphones in meetings or classes:
Let X denote the number of smartphone users who use their smartphones in meetings or classes and X follows Binomial distribution with number of trials $$\displaystyle{n}={11}$$ and probability of success $$\displaystyle{p}={0.55}$$.
Probability mass function of Binomial variable is given by the formula:
$$\displaystyle{P}{\left({X}={x}\right)}=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{\left({1}-{p}\right)}^{{{n}-{x}}}$$
Step 2
Required probability is obtained as follows:
$$P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$$
$$\displaystyle=^{{{11}}}{C}_{{{0}}}{\left({0.55}\right)}^{{{0}}}{\left({1}-{0.55}\right)}^{{{11}-{0}}}+^{{{11}}}{C}_{{{1}}}{\left({0.55}\right)}^{{{1}}}{\left({1}-{0.55}\right)}^{{{11}-{1}}}+^{{{11}}}{C}_{{{2}}}{\left({0.55}\right)}^{{{2}}}{\left({1}-{0.55}\right)}^{{{11}-{2}}}+^{{{11}}}{C}_{{{3}}}{\left({0.55}\right)}^{{{3}}}{\left({1}-{0.55}\right)}^{{{11}-{3}}}$$
$$\displaystyle={1}\times{1}\times{0.0001532278301}+{11}\times{0.55}\times{0.0003405062892}+{55}\times{0.3025}\times{0.0007566806426}+{165}\times{0.166375}\times{0.001681512539}$$
$$=0.0001532278301+0.00206006305+0.01258927419+0.04616067203$$
$$\displaystyle={0.0609632371}$$
$$\displaystyle\approx{0.0610}$$
Thus, the probability that fewer than 4 of them use their smartphones in meetings or classes is 0.0610

###### Not exactly what you’re looking for?
sonSnubsreose6v
This is a binomial with $$\displaystyle{n}={7}\ {p}={0.55}$$
a least 4 of them use their smartphones that is 1-prob(3) of them use them.
$$\displaystyle={0.6083}$$ is answer.
probability 4 use them is $$\displaystyle{7}{C}{4}\cdot{0.55}^{{{4}}}\cdot{0.45}^{{{3}}}={0.2918}$$
for 5 it is 0.2140
for 6 it is 0.0872
for 7 it is $$\displaystyle{0.55}^{{{7}}}\ {\quad\text{or}\quad}\ {0.0152}$$
add them to get 0.6082 (rounding error)
nick1337

The random variable and its distribution is shown below:
From the information, the variable x is the number of adults using smart phones.
From the given information, a random sample of adults with smart phones is considered.
From the given information, it can be observed that there are two possible outcomes ‘using smart phones in meetings’ and ‘using smart phones in classes’. This implies that the first requirement for the binomial distribution is satisfied.
Also, 11 adult Smartphone users are randomly selected is considered. That is, the number of trials in the given experiment is known in advance. This implies that the second requirement for the binomial distribution is satisfied.
The trails (adult) are independent because each variable is independent of the others. This implies that the third requirement for the binomial distribution is satisfied.
From the information given, school figures that 55% use them in meetings or classes that are the probability of success which is same for throughout the experiment. This implies that the fourth requirement for the binomial distribution is satisfied.
Therefore, the distribution is binomial distribution with sample size 11 and probability of success 0.55. That is, $$X \sim Binomial(11,0.55)$$
The probability that fewer than 4 of them use their smartphones in meetings or classes is obtained below:
The required probability is,
P(X<4)=[P(X=0)+P(X=1)+P(X=2)+P(X=3)]
$$=[(\begin{array}{c}11\\ 0\end{array})(0.55)^{0}(1-0.55)^{11-0}+(\begin{array}{c}11\\ 1\end{array})(0.55)^{1}(1-0.55)^{11-1}+(\begin{array}{c}11\\ 2\end{array})(0.55)^{2}(1-0.55)^{11-2}+(\begin{array}{c}11\\ 3\end{array})(0.55)^{3}(1-0.55)^{11-3}]$$
=[0.0002+0.0021+0.0126+0.0462]
=0.0611