A=
AP=<m><1><t>V14</t></1></m><t> to </t><m><1><t>V15</t></1></m><t></t>
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ES=Answers may vary.
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H=$.hint$
INST=
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MS=1
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N=
Q=Find the odds in favor of obtaining.<br>\nV13
SA=Answers may vary.
T=P
TF=-1
TL=
TOL=+1E-4
U=NOUNIT
V0=I[3]
V1=S"A vowel when 1 letter is chosen at random from among the 26 letters of the English alphabet."
V10=S"At least 1 V12 when an ordinary coin is tossed twice."
V11=I[1,2]
V12=L[V11:tail,head]
V13=L[V0:V1,V2,V4,V7,V10]
V14=L[V0:5,1,1,3]
V15=L[V0:21,5,3,1]
V16=S"<i>T</i>"
V17=S"<i>H</i>"
V18=L[V3:one,two,three,four, five, six]
V2=S"A V18 in one roll of a single die."
V3=I[1,6]
V4=S"An V6 number in one roll of a single die."
V5=I[1,2]
V6=L[V5:even,odd]
V7=S"Two V9 when an ordinary coin is tossed twice."
V8=I[1,2]
V9=L[V8:tails,heads]
W=<ostep>V0==1::\nThere are <box>5</box> vowels and <box>2l</box> other letters, so\nthe odds in favor of getting a vowel are <box>5</box>\nto <box>2l</box>.\n</ostep>\n<ostep>V0==2::\nThere is 1 favorable outcome and there are\n<box>5</box> unfavorable outcomes, so the odds are 1\nto <box>5</box> in favor of getting a V18. </ostep>\n<ostep>V0==3::\nYou can get V16V16, V16V17, V17V16, V17V17, so there is l\nfavorable outcome and <box>3</box> unfavorable\noutcomes. Thus, the odds in favor of\ngetting 2 V9 are 1 to <box>3</box>.\n</ostep>\n<ostep>V0==4::\nYou can get V16V16, V16V17, V17V16, V17V17, so there is <box>3</box>\nfavorable outcome and <box>1</box> unfavorable\noutcome. Thus, the odds in favor of\ngetting 2 V9 are <box>3</box> to <box>1</box>.\n</ostep>\n
YN=-1
cnum=4
followup=
ilev=0
mcdm=1
mdm=0
mpc=unspecified
mpn=2
mpv=2.0
mwnum=2
ncd=-1
subject=unspecified
varnum=23