1. Suppose A is the event of getting at exactly one ace when dealt 5 cards from an ordinary deck of cards. Suppose B is the event of getting all hearts when dealt 5 cards. Compute the following probabilities. For questions a-d do not simplify the arithmetic computations. For e you need to do the calculation.
d) P(A&B) (you may use Bayes Rule)
e) Are A and B independent?
a) P(A) = comb(4,1)* comb(48,4) / comb(52,5) = 0.2995
b) P(B) = comb(13,5) / comb(52,5) = 4.952e-4
c) P(A|B) by direction computation = comb(1,1)*comb(12,4) / comb(13,5) = comb(12,4)/ comb(13,5). = .3846
d) P(A&B) = P(A,B) Bayes says: P(A,B) = P(A|B)*P(B) so comb(12,4)/comb(52,5) = 1.9046e-4
e) Setting P(A,B) =?= P(A)*P(B) I get, after a lot of simplifications:
51*50*49* =?= 47*46*45 So they are close but not equal.
Hence not independent. Alternatively, one might just compute the true values for each side.
By calculation P(A)*P(B) = 1.483e-4 so this shows not equal, but I make fewer errors when I don’t calculate.
From the text (new edition) pg 489, do problem 13.6 Beware of difference of bold and non-bold B.
All of these questions can be answered by direct examination of the joint probability table plus the definitions .
P(toothache) = .108+.012+.016+.064 =.2
P(cavity) = <.108+.012+.072+.008, .016+.064+ .144+.576> = <.2 .8>
P(Toothache |cavity) = P(Toothache and cavity )/ P(cavity) =
< .12/.2, .08/.2> = < .6, .4>
P(Cavity|toothache or catch) = P(Cavity and(toothache or catch)) |P(toothache or catch)
P(toothache or catch) = 1 – P(not toothache and not catch) = 1- .008 -.576 = .426 (alternatively add the 6 appropriate 6 entries).
P(Cavity and(toothache or catch) ) = < .108+.012+.072, .016+.064+.144> = < .192, .224>
P(Cavity|toothache or catch) = < .192/.426, .224/.426> = <.4507, 5258>
From the text (new edition) pg489, do problem 13.8
P( test_positive | have_disease ) = .99
P (test_neg | not have_disease) = .99
P( disease) = .0001.
You test positive.
Prob(you have disease|test Positive) = P(test+ | have disease)*P(disease)/ P(test+)
P(test+) = P(test+| disease)*P(disease) + P(test+ | not disease)* P(not disease)
P(test+) = .99*.0001 + .01 * .9999
Plug and chug. = 0.0098 So it is still unlikely that you have the disease.