Probability Questions:
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 ad do not simplify the arithmetic computations. For e you need to do the calculation.
a) P(A)
b) P(B)
c) P(AB)
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) = 4.952e4
c) P(AB) 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(AB)*P(B) so comb(12,4)/comb(52,5) = 1.9046e4
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) = 1.483e4 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 nonbold 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(Cavitytoothache 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(Cavitytoothache 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 diseasetest 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. 