Question

Please help

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?


A.63/3450


B.5/18


C.6/11


D.1/48

Answer 1

B would be your best choice.

Answer 2

@kelsey24 can you explain?

Answer 3

The probability of getting a blue marble is,
\[P(blue)=\frac{7}{25}\]
Now, you have to drop a ball from the 25. So in your bag now there are 24 balls. The probability of getting a yellow marble after the first blue extraction,
\[P(yellow|\text{first blue})=\frac{9}{24}\]
Now, your bag has 23 balls. The probability of getting a red marble if the first was blue and the second yellow,
\[P(yellow|\text{first blue & second yellow})=\frac{4}{23}\]
So the total probability of getting all of this at one time (equivalent to obtain it in three consecutive extractions without replacement),
\[P=\frac{7}{25}\frac{9}{24}\frac{4}{23}=\frac{63}{3450}\]

Answer 4

@john_ES I simplify it and I got 21/1150

Answer 5

Perfect.