Conditional Probability

[rndragster]

I have one more and I am way not sure how to do it, please help!

The table below shows the drink preferences for people in 3 different age groups. If one of the 255 subjects is randomly chosen, what is the probability that the person drinks water given they are under 21. Round your answer to 3 decimal places.

Water Orange juice Cola
Under 21 years 40 25 20
21 – 40 years 35 20 30
Over 40 years 20 30 35


Not sure if it is 40/95 = 0.421 and if so, how do you set up the problem? and if not please help me!

 

[soroban]

Hello, rndragster!

The table below shows the drink preferences for people in 3 different age groups.
If one of the 255 subjects is randomly chosen, what is the probability that the person drinks water, given they are under 21?
Round your answer to 3 decimal places.

. . $\displaystyle \begin{array}{|c||c|c|c||c|} \text{Age} & \text{Water} & \text{O.J,} & \text{Cola} & \text{Total} \\ \hline \hline \text{under 21} & 40 & 25 & 20 & 85 \\ \hline \text{21 - 40} & 35 & 20 & 30 & 85 \\ \hline \text{over 40} & 20 & 30 & 35 & 85 \\ \hline \hline \text{Total} & 95 & 75 & 85 & 255 \\ \hline \end{array}$

$\displaystyle \text{Bayes' Theorem: }\;P(\text{water}\,|\,\text{under 21}) \;=\;\frac{P(\text{water}\,\wedge\,\text{under 21})}{P(\text{under 21})}$


\(\displaystyle \text{The numerator is: }\(\text{water}\,\wedge\,\text{under 21}) \:=\:\frac{40}{255}\)

\(\displaystyle \text{The denominator is: }\(\text{under 21}) \:=\:\frac{85}{255}\)


$\displaystyle \text{Therefore: }\;P(\text{water}\,|\,\text{under 21}) \;=\;\dfrac{\frac{40}{255}}{\frac{85}{255}} \;=\;\frac{40}{85} \;=\;\frac{8}{17} \;\approx\;0.471$

 

[rndragster]

Thank you! I will need to look at this more, and look in my book a little deeper. My professor did not review Bayes's Theorem.