Question

It is observed that 80% of television viewers watch 'NEWS' what is the probability that as least 80% of the viewers in a random sample of five watch NEWS?

Answer 1

By condition, the probability that one viewer watching 'NEWS' is $p = 0.8 \Rightarrow q = 1 - p = 0.2$ .

If at least 80% of the five viewers are watching 'NEWS' , then 4 or 5 viewers are watching  'NEWS'.

Using Bernoulli's formula, we find the corresponding probabilities:

${P_5}\left( 4 \right) = C_5^4{p^4}q = 5 \cdot {0.8^4} \cdot 0.2 = 0.4096$

${P_5}\left( 5 \right) = {p^5} = {0.8^5} = {\rm{0}}{\rm{.32768}}$

Then the wanted probability is

$P = {P_5}\left( 4 \right) + {P_5}\left( 5 \right) = 0.32768 + 0.4096 = 0.73728$

Answer: 0.73728