# Assume that when adults with smartphones are randomly​ selected, 52​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.

Assume that when adults with smartphones are randomly​ selected, 52​% use them in meetings or classes. If 10 adult smartphone users are randomly​ selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.
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Step 1
Given Information:
52% of adults use their smartphones in meetings or classes. i.e., $p=0.52$
If $n=10$ adult smartphone users are randomly selected, to find the probability that fewer than 5 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 $n=10$ and probability of success $p=0.52$.
Probability mass function of Binomial variable is given by the formula:
$P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}$
Step 2
Required probability is obtained as follows:
$P\left(X<5\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)$
${=}^{10}{C}_{0}{\left(0.52\right)}^{0}{\left(1-0.52\right)}^{10-0}{+}^{10}{C}_{1}{\left(0.52\right)}^{1}{\left(1-0.52\right)}^{10-1}{+}^{10}{C}_{2}{\left(0.52\right)}^{2}{\left(1-0.52\right)}^{10-2}{+}^{10}{C}_{3}{\left(0.52\right)}^{3}{\left(1-0.52\right)}^{10-3}{+}^{10}{C}_{4}\left(0.52{\right)}^{4}\left(1-0.52{\right)}^{10-4}$
$=1×1×0.00065+10×0.52×0.0014+45×0.2704×0.0028+120×0.141×0.0059+210×0.73×0.012231$
$=0.0065+0.0728+0.341+0.99828+1.875$
$\approx 3.28773$
Thus, the probability that fewer than 4 of them use their smartphones in meetings or classes is 3.28773