Simple probability 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!
 A: Let
$\ \ \ \ A$ be the event he studies the night before,
$\ \ \ \ B$ be the event he parties the night before.
and
$\ \ \ \ C$ 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...
A: You didn’t round it off correctly, but it appears that you probably made the right calculation: the desired probability is $\frac{0.5}{0.99+0.5}=\frac{50}{149}\approx 0.33557$.
A: 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
A: 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.)
