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A student is going to sit for an exam on Calculus. If he studies the night before the exam, he is got 0.99 probability to pass it.

But, instead if he choose to go to his friend's party , he is got 0.5 probability to pass it.

In order to decide what to do , he tosses a coin.

Finally , the next day he succesfully passes the exam. What is the probability that he gone to the party ?

I used conditional probability and i am getting 0.335

Is it right ?

Thanks in advance!

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I'm getting about .336. I think you're method is right but you forgot to round the last decimal place up. – Brett Frankel Mar 30 '12 at 18:23
Thanks for all the answers people, consider this problem sovled. Your methods are very close to mine, so i am very happy! Cheers! Take care – Adrian Carter Mar 31 '12 at 13:21
you are a lifesaver....i had the same problem....thanks Adrian Carter. Credits from Creta – user28037 Mar 31 '12 at 19:55


$\ \ \ \ A$ be the event he studies the night before,

$\ \ \ \ B$ be the event he parties the night before.


$\ \ \ \ C$ be the event he passes the test.

Then $$\textstyle P(B\mid C)={P(B\cap C)\over P(C)}= { P(C| B)P(B)\over P(C| B)P(B)+P(C|A)P(A)} ={(1/2) (1/2)\over (1/2) (1/2) +(1/2)(.99)} ={1\over 1+(2)(.99)}\approx0.33557047. $$

So, yes...

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Thanks but do you calculate th chance for the coin at all ? I am getting the same result btw – Adrian Carter Mar 30 '12 at 18:30
@Adrian Yes. $P(B)=P(C)={1\over2}$. – David Mitra Mar 30 '12 at 19:42

You didn’t round it off correctly, but it appears that you probably made the right calculation: the desired probability is $\frac{0.5}{0.99+0.5}=\frac{50}{149}\approx 0.33557$.

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you will need Bayes theorem.

what i usually do for conditional probability is make a chart.

|        | pass | !pass |
| party  | .5   | .5    |
| !party | .99  | .01   |

now you can look across the row to see the probability you need. for example. bayes theorem states:

for some event space ${A_i}$,

$$ P(A_i|B) = \frac{P(B|A_j)P(A_i)}{\sum\limits_{j}{P(B|A_j)P(A_j)}} $$ with $ P(B) = \sum\limits_{j}P(B|A_j)P(A_j) $

let's let:

$ A_1 = $ probability of studying

$A_2 = $ probability of partying

$ B_1 $ = probability of passing

(and) for the sake of completion, $B_2$ = probability of not passing

so to find the probability of partying and still passing we use :

$ P(A_2|B_1) = \frac{P(B_1|A_2)P(A_2)}{P(B_1|A_1)P(A_1)+P(B_1|A_2)P(A_2)}$

(to use the table for $P(B_1|A_1)$ or the probability of passing given studying, simply look up the probability of him studying (in this case !partying) and look up the probabilty of him passing which is .99)

as everyone has already stated, the answer is in fact .3556

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Multiply the prior by the likelihood function, then normalize, to get the posterior. Don't worry about the normalizing constant until the last step: $$ \underbrace{\left(\frac 1 2, \frac 1 2 \right)}_\text{prior distribution} \cdot \underbrace{\left(0.99, 0.5\right)}_\text{likelihood} \equiv (1,1)\cdot(99,50)=(99,50)\equiv\underbrace{\left( \frac{99}{149},\frac{50}{149} \right)}_\text{posterior distribution} $$

(The binary relation "$\equiv$" just means two vectors are scalar multiples of each other.)

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wrong post plz delete – Adrian Carter Mar 31 '12 at 13:20
@AdrianCarter : I find your comment cryptic. What I wrote above answers your question. This is a somewhat typical example of Bayes' formula. – Michael Hardy Mar 31 '12 at 23:38

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