1) #4.2(Page 154)
(We can use tree diagram to get it)
Sample space = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
a)All heads: {HHH}
b) A head on the second toss: {HHH, HHT, THH, THT}
c) Exactly 2 tails: {HTT,THT,TTH}
2) #4.8(Page 154)
Two cards are selected in succession from an ordinary deck of cards.
C = clubs, D = diamonds, H = hearts, and S = spades.
List the elements of the events consisting of
a) Both spades: {SS}
b) Either both spades or both clubs: {SS, CC}
c) At least one heart: {CH, DH, HH, SH, HC, HD, HS}
d) A diamond as the second card: {CD, DD, HD, SD}
3) #4.11(Page 158)
Find the probability of each of following events:
Notice that we have 36 possible outcomes totally by Rolling 2 dice.
a) A: the sum of the two dice is less than 5
A = {(1,3); (3,1); (2,2); (1,2); (2,1); (1,1)}
P (A) = n(A)/n(S) = 6/36 = 1/6
b) B: get one or two 6s.
B = {(1,6); (6,1); (2,6); (6,2); (3,6); (6,3); (4,6); (6,4); (5,6); (6,5); (6,6)}
P (B) = n(B)/n(S) = 11/36
c) C: Neither is a 6.
Its complement is: (6,6)
P ( C ) = 1- (1/36) = 35/36
d) D: the sum is 7 or 11
D = { (1,6); (6,1); (3,4); (4,3); (2,5); (5,2); (5,6) (6,5)}
P (D) = n(D)/n(S) =8 /36 = 2/9
4) #4.14(Page 159)
a) List the elements of a sample space S.
S = {EEEE, EEEO, EEOE, EOEE, OEEE, EEOO, EOEO, OEEO, EOOE, OEOE, OOEE, EOOO, OEOO, OOEO, OOOE, OOOO}
b) B: all evens
B = {EEEE}
P (B) = n(B)/n(S) = 1/16
c) C: one or more evens
Its complement is: {OOOO}
P (C) = 1-(1/16) = 15/16
d) D: two or more evens occurring in succession.
D = {EEEE EEEO, EEOE, EOEE, OEEE, EEOO, OEEO, OOEE}
P (D) = n(D)/n(S) = 8/16 = 1/2
5) #4.16(Page 159)
n(S) = 5,245,786
a) P (not winning anything) = (1,947,792 + 2,261,952 + 883,575)/5,245,786
= 727617 / 749398
= 0.97
b) P (getting one or more matches) = 1- (1947792 /5245786)
= 1- 0.37
= 0.63
c) P (winning a free ticket) = 142,800/5,245,786 = 0.027
d) P (getting either $75 or $1500) = (9450 + 216)/5,245,786 = 0.018
e) P (winning megabucks) = 1/5,245,786
6) #4.18(Page 160)
By fundamental principle of counting,
a) number of possible routes in traveling from A to B to C: 4´5= 20
b) number of possible routes in traveling from A to B to C to D: 4´5´8=160
7) #4.23(Page 160)
n(S) = 6´6´6´2´2 = 864
a) P(three 6s and two heads) = 1/864
b) P(three 6s) = (1´1´1´2´2
c) P(two heads) = (6´6´6´1´1) /864 = 1/4