Question

A jar contains 3 red, 2 white, and 1 green marble. Aside from color, the marbles are indistinguishable. Two marbles are drawn at random without replacement from the jar. Let X represent the number of red marbles drawn.

What is the probability of picking 4 marbles and getting one of each color?

Answer 1

0.6

Explanation:

A permutation is an arrangement of outcomes whereas a combination is a grouping of outcomes.

Order matters in case of permutations but does not matter in combinations.

Given:

Number of red marbles = 3

Number of white marbles = 2

Number of green marbles = 1

To find: probability of picking 4 marbles and getting one of each color

Solution:

Total number of outcomes =
`6_C__4` =
`(6!)/(4!\,2!)=(6* 5* 4!)/(4!* 2)=15`

Number of favourable outcomes = 2 red, 1 white and 1 green marble + 1 red, 2 white and 1 green marble =
`3_C_2\,2_C_1\,1_C_1+3_C_1\,2_C_2\,1_C_1` =
`(3!)/(2!\,1!)* (2!)/(1!\,1!)* 1+(3!)/(2!\,1!)* 1* 1=6+3=9`

So, probability of picking 4 marbles and getting one of each color = Number of favourable outcomes/Total number of outcomes =
`(9)/(15)=(3)/(5)=0.6`