Statistics Homework (Need help)?

a)
In Table 1 below, the red numbers show the totals of two dice.
Number of total without restriction = 6 × 6 = 36
The 12 totals less than 5 or greater than 9 are circled.

P(The total of the two dice is less than 5 or greater than 9)
= 12/36
= 1/3

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b)
In table 2 below, the blue numbers show the sums of two dice.
The 20 sums less than 7 or divisible by 3 are circles.

P(The sum of the two dice is less than 7 or is divisible by 3)
= 20/36
= 5/9

a) P(< 5) = 6/36 and P(> 9) = 6/36

so, P(< 5 or > 9) = 12/36 = 1/3

b) P(< 7 or d3) = P(< 7) + P(d3) - P(< 7 and d3)

so, 15/36 + 12/36 - 7/36

i.e. 20/36 = 5/9

:)>

2 = 1 + 1  /// only 1 possibility
3 = 1 + 2  or  2 + 1  ///  2 possibilities
4 = 1 + 3  or  2 + 2  or  3 + 1  /// 3 possibilities
etc

a. less than 5  =  2, 3, or 4  =  1 + 2 + 3 = 6 possibilities
   greater than 9 = 10, 11, or 12  =  3 + 2 + 1 = 6 possibilities

   12/ 36  =  1/3  <<< answer

b.  less than 7  =  1 + 2 + 3 + 4 + 5  =  15 possibilities
   divisible by 3  =  3, 6, 9, 12 ...  but 3 has already been used
   =  5 +  4 + 1 = 10
  total poss  =  25

   25/36  >>>  answer

Look at this: https://useruploads.socratic.org/35BwP7WhTFqPW4FF8jW9_Probability%20-%20Outcomes%20of%202%20die.jpg

There are 36 possibilities.

There are 6 totals less than five (2, 3, 3, 4, 4 and 4).

There are 6 totals greater  than nine (10, 10, 10, 11, 11 and 12).

So there are 6+6=12 possibilities which are "less than 5 or greater then 9".

Probability =12/36 = 1/3

You should now be able to do b) for yourself.

the allowed results are 2 thru 4 and 10 thru 12
combinations that give you that are
11, 12, 21, 22
55, 64, 46, 65, 56, 66

there are 6x6 = 36 possible results altogether

probability is 10/36 or 5/18