A bag contains 88 red, 66 orange, and 99 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 1212 jellybeans such that the number of red ones is 22, the number of orange ones is 44, and the number of green ones is 66? Express your answer as a fraction or a decimal number rounded to four decimal places.

P ( 2 R , 4 O , 6 G ) = 0.0261

Step-by-step explanation:

Given:

- Red beans = 8

- Orange beans = 6

- Green beans = 9

Find:

What is the probability of reaching into the bag and randomly withdrawing 12 jellybeans such that the number of red ones is 2, the number of orange ones is 4, and the number of green ones is 6?

Solution:

- The question requires the number of selection of 12 jellybeans we can make from total available such that out of those 12 we choose 2 Red, 4 Orange and 6 Green.

- For selection we will use the combinations. So to choose 2 Red from 8; Choose 4 Orange from 6 and 6 green from 9 available. The number of possible outcomes with such condition is:

          Outcomes ( 2 R , 4 O , 6 G ) = 8C2 * 6C4 * 9C6

                                                         = 28*15*84

                                                         = 35280

- The total number of outcomes if we randomly select 12 beans irrespective how many of each color we select from available 23 we have:

          Outcomes ( Select 12 from 23 ) = 23C12

                                                               = 1352078

- Hence, the probability for the case is given by:

P ( 2 R , 4 O , 6 G ) = Outcomes ( 2 R , 4 O , 6 G ) / Outcomes ( Select 12 from 23 )

P ( 2 R , 4 O , 6 G ) = 35280 / 1352078 = 0.0261