Conditional Probability: P(W|H)=29,P(W|Hc)=511,P(H)=12,P(H|W)=P(W|H)=\frac{2}{9},P(W|H^c)=\frac{5}{11},P(H)=\frac{1}{2},P(H|W)=?

I was marked wrong for this question on a probability and statistics quiz, but I can’t figure out how to do it correctly:

P(W|H)=29P(W|H)=\frac{2}{9}
P(W|Hc)=511P(W|H^c)=\frac{5}{11}
P(H)=12P(H)=\frac{1}{2}

What is P(H|W)P(H|W)?

My solution was as follows:

P(W)=P(W|H)⋅P(H)+P(W|Hc)⋅P(Hc)=29⋅12+511⋅12=19+522P(W)=P(W|H)\cdot P(H)+P(W|H^c)\cdot P(H^c)=\frac{2}{9}\cdot\frac{1}{2}+\frac{5}{11}\cdot\frac{1}{2}=\frac{1}{9}+\frac{5}{22}
P(W|H)=P(W∩H)P(H)P(W|H)=\frac{P(W\cap H)}{P(H)}
P(W∩H)=P(W|H)⋅P(H)=29⋅12=19P(W\cap H)=P(W|H)\cdot P(H)=\frac{2}{9}\cdot\frac{1}{2}=\frac{1}{9}

Hence P(H|W)=P(H∩W)P(W)=1919+522P(H|W)=\frac{P(H\cap W)}{P(W)}=\frac{\frac{1}{9}}{\frac{1}{9}+\frac{5}{22}}

What is my mistake? I’ve been trying to figure it out but I am stuck.

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Clearly you’ve calculated P(W)P(W) to be 29+522\frac{2}{9}+\frac{5}{22}, but then used 19+522\frac{1}{9}+\frac{5}{22} instead.
– barak manos
2 days ago

  

 

That was just a typo, I meant to write 1/9 the first time too.
– Conor James Thomas Warford Hen
2 days ago

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1 Answer
1

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I am convinced that you did not make any mistakes.

An equality that apart from Pr(H∣W)\Pr(H\mid W) only contains known probabilities: Pr(W∣H)Pr(H)=Pr(H∣W)[Pr(W∣H)Pr(H)+Pr(W∣Hc)(1−Pr(H))]\Pr\left(W\mid H\right)\Pr\left(H\right)=\Pr\left(H\mid W\right)\left[\Pr\left(W\mid H\right)\Pr\left(H\right)+\Pr\left(W\mid H^{c}\right)\left(1-\Pr\left(H\right)\right)\right]

Substitution gives:

2912=Pr(H∣W)[2912+511(1−12)]\frac{2}{9}\frac{1}{2}=\Pr\left(H\mid W\right)\left[\frac{2}{9}\frac{1}{2}+\frac{5}{11}\left(1-\frac{1}{2}\right)\right]

Resulting in: Pr(H∣W)=29122912+511(1−12)=1919+522\Pr\left(H\mid W\right)=\frac{\frac{2}{9}\frac{1}{2}}{\frac{2}{9}\frac{1}{2}+\frac{5}{11}\left(1-\frac{1}{2}\right)}=\frac{\frac{1}{9}}{\frac{1}{9}+\frac{5}{22}}