Probabilistice Modelling

. Jane has a Probabilistic Modelling lecture every Thursday but sometimes decides to take a
break from her studies (B=1) with probability 1/4, in which case she misses the lecture. Her
module tutor looks at her coursework results and wonders whether she missed last week’s
lecture (L=1) or not (L=0). When asked whether she missed a lecture Jane may admits
she did (J=1) with probability 1/2 but will definitely say she did not (J=0) if she actually
attended with probability 1. If she attends, Jane sits next to her best friend Sheehan, who
attends all classes. If asked whether Jane was in class Sheehan normally acknowledges that
Jane missed the lecture (S=1) if (L=1), with probability 3/4; but sometimes he feels sorry for
her and says Jane was present (S=0) although she wasn’t, with probability 1/4. He always
confirms (S=0) if Jane attended the lecture (L=0), with probability 1.
The following probabilities are given:
• Clearly, if Jane takes a break (B=1) she will not be attending the lecture (L=1) with
probability 1, but may miss the lecture for other reasons, even when she does not take
a break from studies (B=0) with probability 1/4;
• when asked about Jane’s attendence, Shaheen responds that he is not sure, and may
have seen Jane (S=0) with probability 3/4, but may have not (S=1) with probability
1/4;
• when asked if she attended class last week, Jane confidently says ’yes’ (J=0).
(c) Compute the probability that Jane did not attend the lecture last week P(L=1|S, J=0)?
[10 marks]
(d) Compute the probability that she was on a break from her studies P(B=1|S, J = 0)?
[9 marks]
Note that all probabilities are fractions, it is easier not to convert them to decimals.

So what have you done, what are you stuck with, ...?

I stucked with (c) and (d) solution. how to solve (c) and (d)