Solution:
The probability of an event is expressed as
[tex]Pr(\text{event)}=\frac{\text{Number of desired outcome}}{Number\text{ of possible outcome}}[/tex]
Given the survey as shown below:
a) Probability that the chosen TV commercial will last 20 to 39 seconds.
Thus,
[tex]\begin{gathered} Pr(20-39\text{ seconds)=}\frac{frequency\text{ of 20-30 seconds}}{total\text{ frequency}} \\ =\frac{38}{78} \\ =0.4871794872 \\ \therefore Pr(20-39\text{ seconds)}\approx0.487\text{ (3 decimal places)} \end{gathered}[/tex]
Hence, the probability that the chosen TV commercial will last 20 to 39 seconds is 0.487.
b) Probability that the chosen TV commercial will last more than a minute.
Thus,
[tex]\begin{gathered} Pr(\text{more than a minute)=Pr(60+)=}\frac{4}{78} \\ =0.05128205128 \\ \Rightarrow Pr(\text{more than a minute)}\approx0.051\text{ (3 decimal places)} \end{gathered}[/tex]
Hence, the probability that the chosen TV commercial will last more than a minute.is 0.051.
c) Probability that the chosen TV commercial will last between 20 and 59 seconds inclusive.
Thus,
[tex]\begin{gathered} Pr(20\text{ and 59 seconds inclusive)=Pr(20-39) and Pr(40-59)} \\ \text{where} \\ \text{Pr(20-39)=}\frac{38}{78} \\ \text{Pr(40-59)=}\frac{19}{78} \\ \text{thus,} \\ Pr(20\text{ and 59 seconds inclusive)=}\frac{38}{78}\times\frac{19}{78} \\ =\: 0.1186719264 \\ \Rightarrow Pr(20\text{ and 59 seconds inclusive)}\approx0.119\text{ (3 decimal places)} \end{gathered}[/tex]
Hence, the probability that the chosen TV commercial will last between 20 and 59 seconds inclusive is 0.119.
