College Algebra & Trigonometry, 2018a

7.6. PROBABILITY 333
And the probability of drawing no defective balls:
16C 2
= 120
20C 2 190 ≈ 0.6316.
Notice that 6+64+120=190and that 0.0316 + 0.3368 + 0.6316 = 1 That is to say
that the probabilities for all the possible events should add up to 1.
This allows us to compute probabilities based on the probability that an event
won’t happen.
If the probability of an event is P (E), then the probability that the event will not
happen is 1 − P (E).
Example
An urn contains 10 red balls and 15 green balls. If six of the balls are drawn at
random, what is the probability that at least one of them is red?
We can calculate this probability by finding the probability that no red balls are
drawn. The number of ways to draw six green balls is 15 C 6 . The number of ways
to draw six balls from the urn is 25 C 6 . So the probability of drawing six green
balls is:
15C 6
= 5005
25C 6 177, 100 ≈ 0.02826
This means that the probability of drawing at least one red ball is 1 − 0.02826 =
0.97174
We could also find this answer by adding up the possibilities of drawing one red
ball, two red balls and so on, up to six red balls:
10C 1 ∗ 15 C 5
+ 10 C 2 ∗ 15 C 4
+ 10 C 3 ∗ 15 C 3
+ 10 C 4 ∗ 15 C 2
+ 10 C 5 ∗ 15 C 1
+ 10 C 6
25C 6 25C 6 25C 6 25C 6 25C 6 25C 6
=0.1696 + 0.3468 + 0.3083 + 0.1245 + 0.0213 + 0.0012 =
0.9717