Combinations (drawing out a pack of cards)

Q: there are 8 red, 9 green and 6 yellow cards in a pack of cards. Five cards are drwan. Find the probability of obtaining 2 red and 3 green cards if it is known that at least one card is green

I don't get the restriction of the question "if it is known that at least one card is green"

the answer is $\displaystyle P=\frac{8C2\times9C3}{23C5-14C5}$, i don't get the $\displaystyle 23C5-14C5$ part $\displaystyle 23C5$ is the total number of ways of choosing 5 cards from 23, 14 is from 23-9,

1. I don't get the "at least" part and how it ties in with $\displaystyle 23C5-14C5$

Re: Combinations (drawing out a pack of cards)

Quote:

Originally Posted by

**aonin** Q: there are 8 red, 9 green and 6 yellow cards in a pack of cards. Five cards are drwan. Find the probability of obtaining 2 red and 3 green cards if it is known that at least one card is green

I don't get the restriction of the question "if it is known that at least one card is green"

the answer is $\displaystyle P=\frac{8C2\times9C3}{23C5-14C5}$, i don't get the $\displaystyle 23C5-14C5$ part $\displaystyle 23C5$ is the total number of ways of choosing 5 cards from 23, 14 is from 23-9,

1. I don't get the "at least" part and how it ties in with $\displaystyle 23C5-14C5$

I think the question you're being asked is that you already know that the card is green, therefore it doesn't count in the total cards for probability. So you're drawing 4 cards from 22... At least that's what I would assume.

Re: Combinations (drawing out a pack of cards)

The total number of ways of drawing 5 cards is 23C5.

The total number of ways of drawing 5 cards, none of which is green, is 14C5.

So the total number of ways of drawing 5 cards of which at least one of them is green is 23C5-14C5.

Re: Combinations (drawing out a pack of cards)

Quote:

Originally Posted by

**TheBoss** I think the question you're being asked is that you already know that the card is green, therefore it doesn't count in the total cards for probability. So you're drawing 4 cards from 22... At least that's what I would assume.

and you would be wrong.

(there is a clue you can use to determine if you should answer a question or not, it is: If you find your self typing "I think XXXXX is ..." and you can't replace this by "XXXXX is ..". You may ignore this clue if the question has been around for a few days and not been answered by someone who knows.)

CB

Re: Combinations (drawing out a pack of cards)

Quote:

Originally Posted by

**aonin** Q: there are 8 red, 9 green and 6 yellow cards in a pack of cards. Five cards are drwan. Find the probability of obtaining 2 red and 3 green cards if it is known that at least one card is green

I don't get the restriction of the question "if it is known that at least one card is green"

the answer is $\displaystyle P=\frac{8C2\times9C3}{23C5-14C5}$, i don't get the $\displaystyle 23C5-14C5$ part $\displaystyle 23C5$ is the total number of ways of choosing 5 cards from 23, 14 is from 23-9,

1. I don't get the "at least" part and how it ties in with $\displaystyle 23C5-14C5$

Let $\displaystyle N_g$ be the number of greens in the draw and $\displaystyle N_r$ the number of reds.

Bayes' theorem tells us that:

$\displaystyle P(N_r=2,N_g=3|N_g\ge 1)P(N_g\ge 1)=P((N_r=2,N_g=3) \& (N_g\ge 1))\\ \\ \phantom{xxxxxxx}=P(N_r=2,N_g=3)$

so:

$\displaystyle P(N_r=2,N_g=3|N_g\ge 1)=\frac{P(N_r=2,N_g=3)}{P(N_g\ge 1)}$

which you could also deduce from a Venn diagram once you realise that the set of outcomes with $\displaystyle N_g\ge 1$ contains the set of outcomes with $\displaystyle N_r=2,N_g=3$

CB