Question

A bag contains 7 red, 5 orange, and 6 green jellybeans. What is the probability of reaching into the bag and randomly withdrawing 12 jellybeans such that the number of red ones is 6, the number of orange ones is 4, and the number of green ones is 2? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

Answer:

Probability of reaching into the bag and randomly withdrawing 12 jellybeans such that the number of red ones is 6, the number of orange ones is 4, and the number of green ones is 2 is (525/18564) = 0.0283

Step-by-step explanation:

We need to select 6 red jelly beans from 7 red ones, 4 Orange jelly beans from 5 Orange ones and 2 green jelly beans from 6 green ones.

This can be done in the following way

⁷C₆ × ⁵C₄ × ⁶C₂ = 7 × 5 × 15 = 525

But totally, 12 jelly beans can be selected from 18 jelly beans in

¹⁸C₁₂ = 18564

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

525/18564 = 0.0283

Answer 2

Answer:

P(A) = 525/18564 = 25/884 or 0.0283

Step-by-step explanation:

The probability of reaching into the bag and randomly withdrawing 12 jellybeans such that the number of red ones is 6, the number of orange ones is 4, and the number of green ones is 2 can be represented with

P(A).

P(A) = N(S)/N(T)

Where

N(S) = Number of possible selection of the specified proportions of jellybeans

N(T) = Number of possible selections of the total number of jellybeans selected from all the jellybeans

Since in this case order of selection is irrelevant, we apply combination.

Given;

Total Number of red , orange and green jellybeans = 7, 5, 6 respectively

Number of red, orange and green jellybeans to be selected = 6, 4, 2 respectively

Total number of jellybeans = 18

Total number of jellybeans to be selected = 12

N(S) = 7C6 × 5C4 × 6C2 = 7×5×15 = 525

N(T) = 18C12 = 18564

P(A) = 525/18564 = 25/884 or 0.0283