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Hi I need help with the following question, I just can't seem to figure it out and have tried multiple solutions:

The probability that a room at Old College had a refrigerator or a TV (or both) is 0.73.

A general survey of all college rooms across the university has been conducted. The survey showed that 80% of college rooms had a refrigerator or a TV (or both). Old College has 24% of the college rooms in the university. What is the probability that a randomly selected college room which had a refrigerator or a TV (or both) is an Old College room? Give your answer to three decimal places.

I tried using Bayes rule to compute 0.73*0.24/0.73*0.24+0.8*0.76 but the answer is 0.219. Can't figure out where I went wrong

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  • $\begingroup$ Please use MathJax to format questions on this site. $\endgroup$
    – saulspatz
    Commented Apr 11, 2018 at 2:52

3 Answers 3

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Let $$A=R \cup T$$ where $R$ is having a refrigerator and $T$ is having a TV. Let $Old$ be the event it is an Old College room.

$$P(A | Old) = .73$$ $$P(A) = .8$$ $$P(Old) = .24$$

Then $$P(Old | A) = \frac{P(A|Old)P(Old)}{P(A)}$$

You seemed to have used the total law of probability calculating $P(A)=P(A \cap Old)+P(A\cap Old^c)$, but miscalculated the second term $P(A \cap Old^c) = P(A|Old^c)P(Old^c)$. More specifically, you used the incorrect value for $P(A | Old^c)$.

However, $P(A)$ is already given in the problem.

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    $\begingroup$ I definitely overcomplicated the question. I understand now. Thanks for your help! $\endgroup$ Commented Apr 11, 2018 at 3:18
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Let $O$ be the event that the room belongs to Old College Let $R$ be the event that a room has a refrigerator. Let $T$ be the event that a room has a TV

\begin{align} P(O|R \cup T) &= \frac{P(O)P(R \cup T|O)}{P(R \cup T)}\\ &= \frac{P(O)P(R \cup T|O)}{P(R \cup T|O)P(O) + P(R \cup T|O^c)P(O^c)} \end{align}

You have used the wrong number for $P(R \cup T|O^c)$

It suffices to use the formula \begin{align} P(O|R \cup T) &= \frac{P(O)P(R \cup T|O)}{P(R \cup T)}\\ \end{align}

to solve the problem since we are given that $P(R \cup T)=0.8$.

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Let $A$ be the event that the randomly selected room is an Old College room. Let $B$ be the event that the randomly selected room has a refrigerator or a TV (or both).

Your solution attempts to use the formula $$ P(A \mid B) = P(B \mid A) \frac{P(A)}{P(B \mid A) P(A) + \underbrace{P(B \mid A^C)}_{\text{?}} P(A^C)}. $$ However, it is incorrect to assume that $P(B \mid A^C) = .8$. We are only given that $P(B) = .8$.

Here is a correct calculation: $$ P(A \mid B) = P(B \mid A) \frac{P(A)}{P(B)} = .73 \left(\frac{.24}{.8} \right) = .219. $$

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