Question 1179137
make a table as shown below.\


<pre>]
                    junior           senior          total

male                  18               7               25
female                10               5               15

total                 28              12               40

</pre>

there are 28 juniors out of a total of 40 so the probability of a student, selected at random, of being a junior is 28/40.


there are 25 males out of a total of 40 so the probability of a student, selected at random, of being a male is 25/40.


there are 28 juniors out of a total of 40, so the probability of a student, selected at random, of being a junior is 28/40.
there are 15 females out of a total of 40, so the probability of a student, selected at random, of being a female is 15/40.


the total is 28 + 15 = 43 out of 40, but .....
10 of those are both a female and a junior.
as such, they are being double counted, 10 as a junior and 10 as a female.
to avoid the double counting, you have to subtract 10 from the total to get a total of 28 + 15 - 10 = 33 that are either a junior or a female.


the formulas take care of this.


let a = the probability of being a junior.
let b = the probability of being a female.


the formula states that p(a or b) = p(a) + p(b) - p(a and b)


that becomes 28/40 + 15/40 - 10/40 = (28 + 15 - 10) / 40 = 33/40.