Question 1153151: If one person is randomly selected, what is the probability of selecting a
person who is Group O or Rh+ ?
| O | A | B |AB |
---------------------------
Type Rh+ |37 |36 | 9 | 3 |
Type Rh- | 7 | 6 | 1 | 1 |
Answer by Edwin McCravy(20060) (Show Source):
You can put this solution on YOUR website! If one person is randomly selected, what is the probability of selecting a
person who is Group O or Rh+ ?
When you have a table like this given
| O | A | B |AB |
---------------------------
Type Rh+ |37 |36 | 9 | 3 |
Type Rh- | 7 | 6 | 1 | 1 |
Always add on a column of totals on the right and
a row of totals on the bottom, like this, and add
the numbers across and down, and the total of the
new column should equal the total of the new row.
| O | A | B |AB |Totals|
---------------------------------|
Type Rh+ |37 |36 | 9 | 3 | 85 |
Type Rh- | 7 | 6 | 1 | 1 | 15 |
----------------------------------
Totals |44 |42 |10 | 4 | 100 |
So we add all those that are either in the "Type Rh+" row or
in the "O" column. That is, the red ones below fit the property
of being in one OR the other category:
| O | A | B |AB |Totals|
---------------------------------|
Type Rh+ |37 |36 | 9 | 3 | 85 |
Type Rh- | 7 | 6 | 1 | 1 | 15 |
----------------------------------
Totals |44 |42 |10 | 4 | 100 |
That's 37 + 36 + 9 + 3 + 7 = 92
It's that many, 92, out of 100, so the answer is
[Notice that we can't just add the "Rh+" row, 85, to the "O" column to get 44+85,
which is too many, 129, because that would be adding the 37 twice, unless we
then subtracted the 37 and then got 129-37 which is 92.]
Edwin
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