Probability Question (1 Viewer)

[laters]

Hi, so I'm stuck on this question...

There are 8 red, 9 green and 6 yellow cards in a pack of cards. Five cards are drawn. Find the probability of obtaining 2R, 3G if it is known that at least 1 card is green.

My answer was (8C2 x 8C2)/(22C4) (I assumed one was already green)

The solution gave a different answer (different when plugged into calculator as well)... (9C3 x 8C2)/(23C5 - 14C5)

I see what they're getting at, but can someone please explain why my answer is numerically different to theirs?

Also, since some of the cards we are choosing are identical, do I need to account for this? If so, please explain how.

Your help is appreciated

 

[braintic]

Looking at 'their' solution, the question has not been specified well enough.
Their solution assumes that cards of the same colour are otherwise distinguishable, and that is not evident from the wording.
I would have interpreted the wording as the cards being distinguished only by colour.

Given their interpretation of the question, your issue is in assuming that a PARTICULAR green card has been included, rather than just ANY green card.

Where is this question from? If it is from a past exam, I would hope the students weren't penalised for interpreting the question the way it reads.

 

[laters]

Looking at 'their' solution, the question has not been specified well enough.
Their solution assumes that cards of the same colour are otherwise distinguishable, and that is not evident from the wording.
I would have interpreted the wording as the cards being distinguished only by colour.

Given their interpretation of the question, your issue is in assuming that a PARTICULAR green card has been included, rather than just ANY green card.

Where is this question from? If it is from a past exam, I would hope the students weren't penalised for interpreting the question the way it reads.

idk if the link works but https://9eeba4054ee764a743f35bb25b5.../3U/James Ruse 2009 3U Trials & Solutions.pdf

It's the 2009 3U Ruse paper, q5

 

[braintic]

Looking at 'their' solution, the question has not been specified well enough.
Their solution assumes that cards of the same colour are otherwise distinguishable, and that is not evident from the wording.
I would have interpreted the wording as the cards being distinguished only by colour.

Given their interpretation of the question, your issue is in assuming that a PARTICULAR green card has been included, rather than just ANY green card.

Where is this question from? If it is from a past exam, I would hope the students weren't penalised for interpreting the question the way it reads.

The link works. I've never been a fan of Ruse papers. I would hope that the smartest Ruse students would have challenged this interpretation of the question back in 2009. To me, this should be interpreted in the same way as having 8 5c pieces, 9 10c pieces, and 6 20c pieces, and asking "if we randomly chose 5 coins, what is the probability of choosing 2 5c pieces and 3 10c pieces, given that you choose at least one 10c piece". Their answer doesn't work for that interpretation.

 

[Ekman]

Hi, so I'm stuck on this question...

There are 8 red, 9 green and 6 yellow cards in a pack of cards. Five cards are drawn. Find the probability of obtaining 2R, 3G if it is known that at least 1 card is green.

My answer was (8C2 x 8C2)/(22C4) (I assumed one was already green)

The solution gave a different answer (different when plugged into calculator as well)... (9C3 x 8C2)/(23C5 - 14C3)

I see what they're getting at, but can someone please explain why my answer is numerically different to theirs?

Also, since some of the cards we are choosing are identical, do I need to account for this? If so, please explain how.

Your help is appreciated

Can you please correct the 14C3 to 14C5, I was going crazy for the past two days trying to think how they would of gotten 14C3...

And as braintic stated, the question assumes that the cards are not identical which should be stated in the question

 

[Kaido]

Can you please correct the 14C3 to 14C5, I was going crazy for the past two days trying to think how they would of gotten 14C3...

And as braintic stated, the question assumes that the cards are not identical which should be stated in the question

Mathlife

 

[braintic]

Can you please correct the 14C3 to 14C5, I was going crazy for the past two days trying to think how they would of gotten 14C3...

And as braintic stated, the question assumes that the cards are not identical which should be stated in the question

Mathlife

But no Englishlife