An online poll asked people how likely they were to see a new movie that was coming out in the summer.Here's how they responded: $$\begin{array} { l | l }
\text { Response } & \text { Number } \
\hline \text { Will definitely see it } & 223 \
\text { Will probably see it } & 251 \
\text { Will probably not see it } & 330 \
\text { Will definitely not see it } & 385 \
\hline \text { Total } & 1189
\end{array}$$ Let's call someone who responded that they would definitely or probably see it a "likely viewer" and the other two categories,"unlikely viewer." If we select two people at random from this sample,what is the probability that one is a likely viewer and one isn't?
A)0.159
B)0.305
C)0.240
D)0.152
E)0.479

Question

An online poll asked people how likely they were to see a new movie that was coming out in the summer.Here's how they responded: $$\begin{array} { l | l } 
\text { Response } & \text { Number } \
\hline \text { Will definitely see it } & 223 \
\text { Will probably see it } & 251 \
\text { Will probably not see it } & 330 \
\text { Will definitely not see it } & 385 \
\hline \text { Total } & 1189
\end{array}$$ Let's call someone who responded that they would definitely or probably see it a "likely viewer" and the other two categories,"unlikely viewer." If we select two people at random from this sample,what is the probability that one is a likely viewer and one isn't?
A)0.159
B)0.305
C)0.240
D)0.152
E)0.479

Quizplus · Accepted Answer

The number of likely viewers is the sum of those who will definitely see it and those who will probably see it, which is $223 + 251 = 474$. The number of unlikely viewers is the sum of those who will probably not see it and those who will definitely not see it, which is $330 + 385 = 715$. The total number of ways to choose 2 people out of 1189 is $\binom{1189}{2}$. The probability of choosing one likely viewer and one unlikely viewer is the sum of the probability of choosing a likely viewer first and an unlikely viewer second and the probability of choosing an unlikely viewer first and a likely viewer second. This can be calculated as $\frac{474}{1189} \times \frac{715}{1188} + \frac{715}{1189} \times \frac{474}{1188}$. Simplifying this gives a probability of approximately 0.479, making E the correct answer.

An Online Poll Asked People How Likely They Were to See