Question 1160922
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A student walks to school everyday. 
If it is not raining, the probability that he/she is not late is four times the probability that he/she is late. 
If it is raining, the probability that he/she is late is twice the probability that she/he is not late. 
On any particular day, it is three times as likely not to rain as it is raining. 
If in one particular day the student is late, what is the probability that it was raining on that day?
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I edited your post to make it grammatically and semantically correct.



<pre>
            To solve this tricky problem, we should accurately formulate 
                    what is given and what we want to find.


                           +-----------------+
                           |  What is given  |
                           +-----------------+


From the problem, we are given


    (a) if it is not raining, P(not late) = 4*P(late) and P(not late) + P(late) = 1.

        It implies that if it is not raining, P(not late) = 4/5, P(late) = 1/5.



    (b) if it is raining, P(late) = 2*P(not late) and P(not late) + P(late) = 1.

        It implies that if it is raining, P(late) = 2/3, P(not late) = 1/3.



    (c) On a particular day, P(not to rain) = 3*P(rains) and P(not to rain) + P(rain) = 1.

        It implies that on a particular day P(not to rain) = 3/4,  P(rains) = 1/4.



Let's write (a) and (b) in compact form, using the symbol of conditional probabilities.


Then (a) takes form  P(not late | not R) = 4/5;  P(late | not R) = 1/5.

     (b) takes form  P(not late | R)     = 1/3;  P(late | R)     = 2/3.     


                      +------------------------+
                      |  What we want to find  |
                      +------------------------+


      We want to find the conditional probability  P(raining | late).



This conditional probability is the ratio  P(raining AND late)/P(late).     (1)


The denominator in formula (1) is the probability that the student is late 

    P(late) = P(rains)*P(late | R) + P(not to rain)*P(late | not R) = {{{(1/4)*(2/3)}}} + {{{(3/4)*(1/5)}}} = {{{2/12 + 3/20}}} = {{{10/60 + 9/60}}} = {{{19/60}}}.


The probability that in one particular day two events will happen - the student is late AND it raining is  

    P(late AND R) = P(late | R)*P(rains) = {{{(2/3)*(1/4)}}} = {{{2/12}}} = {{{1/6}}}.   


Therefore, the conditional probability  P(late AND raining)/P(late) = {{{((1/6))/((19/60))}}} = {{{60/(6*19)}}} = {{{10/19}}}.


<U>ANSWER</U>.  The conditional probability  P(late AND raining | late) = {{{10/19}}}.
</pre>

Solved.