Question

255 people were asked which evening news do they watch most often. If one of these individuals was randomly chosen. Find the probability that the person watches ABN or ABS

Answer 1

given that the individual is a man...there are 15 men that watch ABN 20 watching ABS 50 watching SBC and 40 are watching other. 125 total. Then there are 60 women watching ABN, 20 ABS , 10 SBC 40 other. Total 130

Answer 2

Yes, I can read it, I'm just not sure exactly what it is asking for. Just the probability that a man is watching?

Answer 3

Are we just looking at the men?

Answer 4

P(ABN or ABS| man)=

Answer 5

Considering the men only, of which there are 125, if 15 watch ABN and 20 watch ABS, the total watching ABN and ABS is 35. So 35/125 watch ABN and ABS

Answer 6

3.57:1 is the ratio.

Answer 7

\[\begin{matrix}
\underline{~~~~~~~~~~}&|&\underline{\text{ABS}}&|&\underline{\text{ABN}}&|&\underline{\text{SBC}}&|&\underline{\text{other}}&|&\underline{\text{total}}\\
\text{man}&|&20&|&15&|&50&|&40&|&125\\
\underline{\text{woman}}&|&\underline{20}&|&\underline{60}&|&\underline{10}&|&\underline{40}&|&\underline{130}\\
\text{total}&|&40&|&75&|&60&|&80&|&255
\end{matrix}\]
\[\begin{align*}P(\text{ABN or ABS}|\text{man})&=\frac{P(\text{(ABN or ABS) and man})}{P(\text{man})}\\
&=\frac{P(\text{(ABN and man) or (ABS and man}))}{P(\text{man})}\\
&=\frac{P(\text{ABN and man})}{P(\text{man})}+\frac{P(\text{ABS and man})}{P(\text{man})}
\end{align*}\]
Watching ABN and watching ABS are mutually exclusive, I'm assuming... For mutually exclusive events \(A\) and \(B\), you have \(P(A\cup B)=P(A)+P(B)\).

So, from the table you know that
\[P(\text{ABN and man})=\frac{15}{255}\\
P(\text{ABS and man})=\frac{20}{255}\\
P(\text{man})=\frac{125}{255}\]