Question

assume that when adults with smartphones are randomly selected, 44% use them in meetings or classes. If 10 adult smartphones users are randomly selected find the probability that fewer than 3 of them

Answer 1

Solution: The given random experiment follows Binomial distribution with $n=10,p=0.44$

Let $X$ be the number of adults who use their smartphones in meetings or classes.

Therefore, we have to find:

$P(X<3)$

We know the binomial model is:

$P(X=x)=\binom{n}{x} p^{x} (1-p)^{n-x}$

$\therefore P(X<3) = P(X=0)+P(X=1) +P(X=2)$

                       $=\binom{10}{0}0.44^{0}(1-0.44)^{12-0}+\binom{10}{1}0.44^{1}(1-0.44)^{10-1}+\binom{10}{2}0.44^{2}(1-0.44)^{10-2}$

                       $=1 \times 1 \times 0.0030 + 10 \times 0.44 \times 0.0054 + 45 \times 0.1936 \times 0.009672$

                       $=0.0030+0.0238+0.0843$

                       $=0.1111$

Therefore, the probability that fewer than 3 of them is 0.1111