Question

News agency would like to conduct a study on the channel and time in which target customers use to watch the news. It was found that 54% like to watch news from online platform, 45% like to watch news in the morning, and 32% like both. If a person in the survey is randomly selected, find the probability that the person neither like to watch the news from online platform nor watch news in the morningInstruction: Answer with 4 decimal places

Answer 1

Answer:

The probability that a randomly selected person neither like to watch the news from online platforms nor watch news in the morning is 0.6700

Expanation:

Parameters:

54% like to watch online platform

$p\mleft(o\mright)=0.54$

45% like to watch news in the morning

$p\mleft(n\mright)=0.45$

32% like to watch both

$p\mleft(o\cup b\mright)=0.32$

The probability that a randomly selected person neither like to watch the news from online platforms nor watch news in the morning is:

$\begin{gathered} p(o\cap n)=p\mleft(o\mright)+p\mleft(n\mright)-p\mleft(o\cup n\mright) \\ =0.54+0.45-0.32 \\ =0.6700 \end{gathered}$