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i have following problem,

"Students who party before an exam are twice as likely to fail the exam as those who don't party (and presumably study). Of 20% of students partied before the exam, what percentage of students who failed the exam went partying?"

i believe that this problem related to conditional probability, but i couldn't find all necessary elements for answer. appreciate your help.

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  • $\begingroup$ Do you understand Bayes’ theorem? $\endgroup$ – gHem Dec 20 '18 at 4:38
  • $\begingroup$ Is that supposed to read If 20% of students... instead of Of 20% of students... ? $\endgroup$ – David Diaz Dec 20 '18 at 4:53
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    $\begingroup$ An alternative approach to Aaron's answer is to first write down the given info as: \begin{align}P[fail \: \mid \: party] &= 2P[ fail \: \mid \: party^c]\\ P[party] &= 0.2\end{align} and you want to compute $P[party \: \mid \: fail]$. $\endgroup$ – Michael Dec 20 '18 at 5:18
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Let $x$ be the total number of students and $p$ be the probability of a student who didn't party failing the exam. The probability of a student who partied before the exam failing the exam is then $2p$.

$x/5$ students partied, out of which $2px/5$ failed. Out of the $4x/5$ who didn't party, $4px/5$ failed the exam. The total students who failed is $2px/5+4px/5=6px/5$, out of which $2px/5$ partied. The required probability is $2/6=1/3$.

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I would recommend drawing a tree diagram to start with. Look at https://www.mathsisfun.com/data/probability-events-conditional.html for more information.

In your case, I assigned percentages making sure to fulfill your condition.

20% partied - Of them, 60% of them failed and 40% of them passed, while
80% no party - Of them, 30% of them failed and 70% of them passed.

Using these numbers, you can that if taking a sample out of $100$ for example, you will see that 12 people failed who went partying and 24 failed who didn't party. So the percentage who failed (and went partying) is $33.3$%.

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