Probability Question

[lilietta2000]

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Can any body help with this question please?

There are 3 red , 4 white and 5 blue balls in a jar. what is the probability of drawing a red a white and a blue ball w/o replacement?

Answer: 3/11

 

[nfg05]

Can any body help with this question please?

There are 3 red , 4 white and 5 blue balls in a jar. what is the probability of drawing a red a white and a blue ball w/o replacement?

Answer: 3/11

Depends on whether it has to happen in that order. Assuming order matters:

Probability of drawing the red ball is 3/12
Probability of drawing the white bal is 4/11 (since the one red ball is now gone)
Probability of drawing the blue ball is 5/10

Now multiply the probabilities to get the overall probability (3/12)(4/11)(1/2) = (12/264) = 0.04545

If the order does not matter, this can happen in 6 different ways:
white red blue
white blue red
red white blue
red blue white
blue red white
blue white red

So the overall probability of it happening, in ANY order is 6*.04545 = about 0.27

Edit: 3/11 is about 0.27, so it looks like the order doesn't matter. Missed that in your post initially.

As an aside, which you may already understand, if the probability of event A happening is X and the probability of event B happening is Y, and the two events are independent, then the probability of event A AND event B happening is X*Y.

 

[HawkeyePostOp]

3 events, order doesn't matter

3! * [(3/12)*(4/11)*5/10] = 3/11

is this from an actual MCAT book?

 

[engineeredout]

Since you're doing a "without replacement" problem, use the hypergeometric distribution.


Statistics major FTW.

 

[lilietta2000]

3 events, order doesn't matter

3! * [(3/12)*(4/11)*5/10] = 3/11

is this from an actual MCAT book?

This is for DAT

 

[lilietta2000]

Depends on whether it has to happened in that order. Assuming order matters:

Probability of drawing the red ball is 3/12
Probability of drawing the white bal is 4/11 (since the one red ball is now gone)
Probability of drawing the blue ball is 5/10

Now multiply the probabilities to get the overall probability (3/12)(4/11)(1/2) = (12/264) = 0.04545

If the order does not matter, this can happen in 6 different ways:
white red blue
white blue red
red white blue
red blue white
blue red white
blue white red

So the overall probability of it happening, in ANY order is 6*.04545 = about 0.27

Edit: 3/11 is about 0.27, so it looks like the order doesn't matter. Missed that in your post initially.

As an aside, which you may already understand, if the probability of event A happening is X and the probability of event B happening is Y, and the two events are independent, then the probability of event A AND event B happening is X*Y.

Thanks!

What would it be like if order mattered?

 

[engineeredout]

Thanks!

What would it be like if order mattered?

Matt said it at the beginning of his post.

 

[RPedigo]

Moved to DAT forum. I think the question has been answered as well. Good luck.