Answer:
A. 1/220 B. 1/22 C. 3/11
Step-by-step explanation:
This question uses combinations -- counting the number of ways a selection can be made from a set of objects (without arranging them after the selection is done).
Notations for combinations:
The number of ways to select r things from a set of n things is
[tex]C(n,r)=\frac{n!}{r!(n-r)!}[/tex] where the exclamation points mean factorial.
[tex]n!=1\cdot2\cdot3\ldots\cdot n[/tex] and 0! is defined to be 1.
There are two other commonly used symbols for this:
[tex]_nC_r[/tex] and [tex]n \choose x[/tex].
In all three parts, the number of ways to choose 3 balls from a set of 12 is
[tex]C(12,3) = \frac{12!}{3!(12-3)!}=\frac{12\cdot 11 \cdot 10 \cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot4\cdot3\cdot2\cdot1}{(3\cdot2\cdot1)(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1)}[/tex]
Notice that 9 factors in the denominator cancel 9 factors in the numerator, leaving
[tex]C(12,3) = \frac{12!}{3!(12-3)!}=\frac{12\cdot 11 \cdot 10}{3\cdot2\cdot1}=220[/tex]
A. The number of ways to choose 3 red balls from the 3 red balls is C(3, 3) = 1, so the probability is 1 / 220.
B) The number of ways to choose 3 green balls from the set of 5 green balls is [tex]C(5,3)=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=\frac{5\cdot 4}{2}=10[/tex] out of 220, so the probability is 10/220 = 1/22.
C) The number of ways to choose 1 red is C(3, 1) = 3. Choose 1 green in C(5, 1) = 5 ways. Choose 1 white in C(4, 1) = 4 ways.
The probability is [tex]\frac{3\cdot 5 \cdot 4}{220}=\frac{60}{220}=\frac{3}{11}[/tex]
