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