Use the following scenario to answer the question: A bag contains 5 green marble

label Algebra
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Use the following scenario to answer the question:

A bag contains 5 green marbles, 4 red marbles, 7 blue marbles, and 9 yellow marbles.
 
If 3 marbles are drawn from the bag one at a time and not replaced, what is the probability of drawing a blue, yellow, and red marble in that particular order?
Jun 26th, 2015

Thank you for the opportunity to help you with your question!

Total number of marbles = 5+4+7+9 = 25

First draw - Probability of selecting a blue marble = 7C1/25C1 = 7/25
Second draw, Probability of selecting a yellow marble = 9C1/24C1 =  9/24 .................)as now total number of marbles is 24)
Third draw, probability of selecting a red marble = 4C1 / 23C1 = 4/23

The probability of all three events to happen together is = (7/25)*(9/24)*(4/23) = 21/1150
Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 26th, 2015

The only answers are 1/48, 5/18, 63/3450 and 6/11

Jun 26th, 2015

Alright, its simple again

The answer I have provided above is 21/1150

now, multiply numerator and denominator by 3

= 21*3 / 1150*3

= 63/3450

so the answer is 63/3450, option 3

Jun 26th, 2015

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Jun 26th, 2015
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