Prob Exercises Solution

Consider the following equations (alternative conditions for independence)



P(A ∧ B) = P(A) × P(B)



P (A|B) = P (A)



show that (i) implies (ii), and also that (ii) implies (i)




Suppose Harry Potter plays for the Gryffindor team, and suppose he ‘often’ catches the so-called Golden Snitch, and also that ‘often’ Gryffindor win. Suppose the numbers actually are as follows





gw
¬gw








ps means ‘Harry Potter caught the Golden Snitch’
ps
28
2
¬ps
140
30
gw means ‘Gryffindor won’











Calculate P (gw | ps), the conditional prob that Gryffindor won given that Harry Potter caught the Golden Snitch and also indicate which counts are irrelevant to this calculation



Calculate P (ps | gw), the conditional prob that Harry Potter caught the Golden Snitch given that Gryffindor won and also indicate which counts are irrelevant to this calculation



A sound clip from YouTube may or may not have been produced by Victor Meldrew. A clip may or may not contain the phrase ‘I don’t believe it’.



You hear a clip coming from YouTube containing the phrase ‘I don’t believe it’ and want to work out the probability that the speaker was Victor Meldrew.




Formalize with 2 discrete variables



discrete Speaker, values in {‘Victor Meldrew’, ‘Other’}



discrete DBI, values in {true, f alse}, for whether a sound clip contains the phrase ‘I don’t believe it’



Let vmel stand for Speaker = ‘Victor Meldrew’, dbi stands for DBI = true




Work out which of vmel or ¬vmel is likelier, given dbi, supposing the probabilities p(vmel) = 0.01, p(dbi|vmel) = 0.95, p(dbi|¬vmel) = 0.01



Do the same assuming p(vmel) = 0.15, p(dbi|vmel) = 0.95, p(dbi|¬vmel) = 0.01



Do the same assuming p(vmel) = 0.01, p(dbi|vmel) = 0.95, p(dbi|¬vmel) = 0.001



Consider someone who lives in a basement flat. Sometimes it is quite noisy in the flat, and some-times not. Sometimes it is rather cool in the flat, and sometimes not. Let noisy be a variable indicating whether it is rather noisy or not, on a given day, and let cool be a variable indicating whether it is rather cool or not.



Consider the frequency table






noisy : +
noisy : −
(1)
cool : +
62
108


cool : −
38
292





find p(cool : +) and p(cool : +|noisy : +)




and conclude from this whether or not cool : + is independent of noisy : +




Unknown to the occupant of the flat there is ventilator fixture in the wall which can be opened and shut to let air from the street outside in, or keep it out. Unknown to the occupant a pet cat plays about with this at night-time, sometimes leaving it open and sometimes leaving it shut. The table



(1) concerns 500 days. The two tables below split these into a group of 100 days where the cat has left the ventilator open (2), and 400 days where the cat has left it shut (3)



open : +


noisy : +
noisy : −
(2)










cool : +


54
36


cool : −


6
4


open : −


noisy : +
noisy : −
(3)












cool : +


8
72


cool : −


32
288















With reference to the table (2), find p(cool : +|open : +) and p(cool : +|open : +, noisy : +)




and conclude from this whether or not cool : + is conditionally independent of noisy : + given




open : +.




With reference to the table (3), find p(cool : +|open : −) and p(cool : +|open : −, noisy : +)




and conclude from this whether or not cool : + is conditionally independent of noisy : + given




open : −







2