Question

Suppose that $80$% of all families own a television set. If $10$ families are interviewed at random, find the probability that
i) seven families own a television set.
ii) at most three families own a television set.

Answer 1

$X=$Number of families who own a television set .

$P=$ Probability of families who own a television set

Given:

$p=\frac{80}{100}=\frac{4}{5}\therefore q=1-p=1-\frac{4}{5}=\frac{1}{5}$

Given ,$n=10$

$\therefore P(X=x)=C_x^{10}{p}^x{q}^{10-x}a\left)P\right(X=7)=C_7^{10}{p}^7{q}^3=C_7^{10}{\left(\frac{4}{5}\right)}^7{\left(\frac{1}{5}\right)}^3=\frac{10\times 9\times 8}{3\times 2}{\left(\frac{4}{5}\right)}^7{\left(\frac{1}{5}\right)}^3$

$=120{\left(\frac{4}{5}\right)}^7{\left(\frac{1}{5}\right)}^3=\frac{120\times 4^7}{5^{10}}=\frac{393,216}{1,953,125}\cong 0.2013$

$b\left)P\right(X\le 3)=C_0^{10}{p}^0{q}^{10}+C_1^{10}p{q}^9+C_2^{10}{p}^2{q}^8+C_3^{10}{p}^3{q}^7={\left(\frac{1}{5}\right)}^{10}+10(\frac{4}{5}){\left(\frac{1}{5}\right)}^9+45{\left(\frac{4}{5}\right)}^2{\left(\frac{1}{5}\right)}^8+120{\left(\frac{1}{5}\right)}^3{\left(\frac{4}{5}\right)}^7={\left(\frac{1}{5}\right)}^{10}+10(\frac{4}{5^{10}})+45(\frac{16}{5^{10}})+120(\frac{4^7}{5^{10}})=\frac{1+40+(45\times 16)+(120\times 4^7)}{5^{10}}=0.2014$