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

Answer:

0.6

Step-by-step 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 = $6_C__4$ = $\frac{6!}{4!\,2!}=\frac{6\times 5\times 4!}{4!\times 2}=15$

Number of favourable outcomes = 2 red, 1 white and 1 green marble + 1 red, 2 white and 1 green marble = $3_C_2\,2_C_1\,1_C_1+3_C_1\,2_C_2\,1_C_1$ = $\frac{3!}{2!\,1!}\times \frac{2!}{1!\,1!}\times 1+\frac{3!}{2!\,1!}\times 1\times 1=6+3=9$

So, probability of picking 4 marbles and getting one of each color = Number of favourable outcomes/Total number of outcomes =$\frac{9}{15}=\frac{3}{5}=0.6$