Question

Suppose that $80\%$ of all families own a television set. If $5$ families are interviewed at random,
find the probability that:
$\left(i\right)$ Three families own a television set.
$\left(ii\right)$ At least two families own a television set.

Answer 1

$X=Nooffamilies$

$P=Probabilityoffamilies$

$P=80percentage=\frac{80}{100}=\frac{4}{5}$

$q=1-p=1-\frac{4}{5}=\frac{1}{5}$

$n=5,X-B(5,\frac{4}{5})$

$P(X=x)=C_x{p}^x{q}^{n-x}=C_x{p}^x{q}^{5-x}$

Families own television set $P=P(X=3)$

$=C_3{\left(\frac{4}{5}\right)}^3{\left(\frac{1}{5}\right)}^{5-3}$

$=\frac{128}{625}$

$=0.205$

Atleast two families $P(X\ge 2)=1-P(X<2)$

$=1-[P(X=0)+P(X=1)]$

$=1-[C_0{\left(\frac{4}{5}\right)}^0{\left(\frac{1}{5}\right)}^5+C_1{\left(\frac{4}{5}\right)}^1\ast {\left(\frac{1}{5}\right)}^5]$

$=1-[\frac{1}{55}+\frac{20}{55}]$

$=1-\frac{21}{3125}$

$=\frac{3104}{3125}$