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Sagot :
assume that you only include whole numbers (1,2,3,4,5,6,7,8,9) and not 3.5 and such
so if 1 is odd and less than 5 then it is
1 or 3, since 5 isn't included
then the other number, to be less than 5 when added,
must be
1+x<5
3+x<5
solve each
1+x<5
subtract 1
x<4
set of answers are 1,2,3
3+x<5
subtract 3
x<2
set of answer is 1
so the possible numbers are
1,2,3
that is 3 numbesr out of 9 so
probability=(total desired outcomes)/(total possible outcomes) so
disred outcomes=3
total possible=9
3/9=1/3
the probabiltiy is 1/3
so if 1 is odd and less than 5 then it is
1 or 3, since 5 isn't included
then the other number, to be less than 5 when added,
must be
1+x<5
3+x<5
solve each
1+x<5
subtract 1
x<4
set of answers are 1,2,3
3+x<5
subtract 3
x<2
set of answer is 1
so the possible numbers are
1,2,3
that is 3 numbesr out of 9 so
probability=(total desired outcomes)/(total possible outcomes) so
disred outcomes=3
total possible=9
3/9=1/3
the probabiltiy is 1/3
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