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Conditional Probability

rndragster

New member
Joined
Jan 25, 2011
Messages
4
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

Elite Member
Joined
Jan 28, 2005
Messages
5,588
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.

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


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


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

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


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

 

rndragster

New member
Joined
Jan 25, 2011
Messages
4
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.
 
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