={HHH, HHT, HTH, THH, HTT, THT, TTH,
TTT}
STAT/MATH 394B
2.4.1. ={HHH, HHT, HTH, THH, HTT, THT, TTH,
TTT}
(a) A={HHH, HHT, HTH, HTT} P(A) = 1/2
B={HHT, HTH, THH, TTT} P(B)=1/2 (we ned to count 0 as an even number)
={HHT,HTH} so 
=1/4=1/2
x 1/2; YES, independent events
(b) A={TTH,TTT} P(A)=1/4; B={HTT,TTT} P(B)=1/4 
={TTT}
so
P(
)=1/8 _1/4 x 1/4; NO, not independent
(c) A={HHH, HHT, TTH, TTT} P(A)= 1/2
B={HHH, THH, HTT, TTT} P(B)=1/2
={HHH,TTT} P(
) = 1/4
= 1/2 x 1/2; YES, independent
2. 4.4 (a)
(b)
(c)
(if A is independent of B then it is independent
of Bc).
(d) P(exactly one) =
(e) k=0: 
from (b)
k=1: 7/20 from (d)
k=3: 1/10 from (a)
k=2: 1-1/5-1/10-7/20 = 29/60
2.4.7 
P(B) = 1 or P(A) = 0.
(Above the infinity sign should be an implication arrow)
2.4.12 (a) 
(b) Using (a), 
(c) Since (P(B|A)<P(B), and 
we see
that
