Question

A bag contains 9 red, 8 orange, and 6 green jelly beans. What is the probability of reaching into the bag and randomly withdrawing 17 jellybeans such that the number of red ones is 7, the number of orange ones is 7, and the number of green ones is 3?

Answer 1

Answer: Pt = 2/81800719 or Pt = 2.4449662844E−8

the probability of reaching into the bag and randomly withdrawing 17 jellybeans such that the number of red ones is 7, the number of orange ones is 7, and the number of green ones is 3 is = 2.4449662844E−8 or 2/81800719

Step-by-step explanation:

Given;

Total number of jelly beans in the bag = 9+8+6 = 23 (9 red, 8 orange and 6 green)

We want to find the probability of randomly withdrawing 17 jelly beans such that the number of red ones is 7, orange ones 7, and green ones 3.(without replacement)

Pt = total possible ways of withdrawing the jelly bean in that particular proportion Tp / total possible ways of withdrawing 17 jelly bean Tt

Pt = Tp/Tt

Tp = 9P7 × 8P7 × 6P3 = 9!/2! × 8!/1! × 6!/3!

Tp = 877879296000

Tt = 23P17 = 23!/6!

Tt = 35905578804006912000

Pt = Tp/Tt

Pt = 877879296000/ 35905578804006912000

Pt = 2/81800719

Pt = 2.4449662844E−8

Answer 2

Answer:

63/12167

Step-by-step explanation:

Probability is given by number of possible outcomes ÷ number of total outcomes

Total number of jellybeans = 9+8+6 = 23

Assuming the withdrawal is done with replacement

Probability (of withdrawing 17 jellybeans with 7 red, 7 orange and 3 green) = 7/23 × 7/23 × 3/23 = 63/12167