Question

A bag contains 3red, 4white, 5green balls. Three balls are selected without replacement. Find the probability that the three balls chosen are:

A) all red
B) all green
C)one of each color

Answer 1

a) 1/220

b) 1/22

c) 3/11

Explanation:

Check the attachment.

Answer 2

A. 1/220 B. 1/22 C. 3/11

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

`C(n,r)=(n!)/(r!(n-r)!)` where the exclamation points mean factorial.

`n!=1\cdot2\cdot3\ldots\cdot n` and 0! is defined to be 1.

There are two other commonly used symbols for this:

`_nC_r` and
`n \choose x`.

In all three parts, the number of ways to choose 3 balls from a set of 12 is

`C(12,3) = (12!)/(3!(12-3)!)=(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))`

Notice that 9 factors in the denominator cancel 9 factors in the numerator, leaving

`C(12,3) = (12!)/(3!(12-3)!)=(12\cdot 11 \cdot 10)/(3\cdot2\cdot1)=220`

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
`C(5,3)=(5!)/(3!(5-3)!)=(5!)/(3!2!)=(5\cdot 4)/(2)=10` 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
`(3\cdot 5 \cdot 4)/(220)=(60)/(220)=(3)/(11)`