The Solution:
Given the table below:
Required:
To find the probability that a student selected at random is a male or got a 'C'.
The required probability is:
[tex]P(a\text{ Male or a C})=P(Male)+P(a\text{ C\rparen}-P(M\cap C)[/tex]
In this case,
[tex]\begin{gathered} P(a\text{ Male})=\frac{41}{80} \\ \\ P(a\text{ C})=\frac{17}{80} \\ \\ P(M\cap C)=\frac{6}{80} \end{gathered}[/tex]
Substituting, we get
[tex]\begin{gathered} P(a\text{ Male or a C})=\frac{41}{80}+\frac{17}{80}-\frac{6}{80}=\frac{41+17-6}{80}=\frac{52}{80}=0.65 \\ \approx0.650 \end{gathered}[/tex]
Therefore, the correct answer is 0.650 (3 decimal place)