Advertising on television

A television station claims about its broadcasting schedule that advertising takes up 76 minutes of broadcasting during the whole day and is broadcast at random intervals.
a) what is the probability that if you turn the television to this station, it will be broadcasting advertising?
b) a second television station has in its broadcasting 13.3% of advertising. Assume that both stations broadcast advertising independently of each other. What is the probability that at one moment both stations broadcast advertising simultaneously?
c) what is the probability that at least one of these channels broadcasts advertising?

Final Answer:

p₁ =  0.0528
p₂ =  0.007
p₃ =  0.1788

Step-by-step explanation:

$$\begin{gathered} t_1=76\text{min} \\ {t}_2=24\cdot 60=1440\text{min} \\ {p}_1=\frac{t_1}{t_2}=\frac{76}{1440}=0.0528 \end{gathered}$$

$$\begin{gathered} r_2=13.3\%=\frac{13.3}{100}=0.133 \\ {p}_2=r_2\cdot p_1=0.133\cdot 0.0528=0.007 \end{gathered}$$

$$\begin{gathered} A\cup B=A+B-A\cap B \\ {p}_3=p_1+r_2-p_2=0.0528+0.133-0.007=0.1788 \end{gathered}$$

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