From a deck of ten yellow, ten blue, ten green and ten red cards, ten cards are chosen at random.

Question

The question: From a deck of ten yellow, ten blue, ten green and ten red cards, ten cards are chosen at random. Find the probability that there is at least one blue card chosen.

Since the question didn't say ten "different" yellow cards, ten "different" red cards and so on, I read this question as a combination with repetition question, so this is what I did:

$$P(\text{at least one blue card})=\frac{{}^4H_9}{{}^4H_{10}}=\frac{{}^{12}C_9}{{}^{13}C_{10}}=\frac{10}{13}$$

But here's what the answer had

$$1-\frac{^{30}C_{10}}{^{40}C_{10}}$$

i.e. they saw the ten yellow cards as ten different yellow cards and so on.

I think I'm in the right, but I am curious as to whether what other people think.

This was a question that my tutor gave me, so I'm disagreeing with my tutor.

Opinion much appreciated.

Answer 1

First of all, you might have mixed up something at the value of $r$ in ${}_nH_r$. This symbol answers "if we have $n$ suits, how many ways we can take $r$ cards from these suits". So it looks like your choice make sense if you follow this process $10$ times: take $1$ card, place it back, and shuffle. However, what you should have calculated would be $${}_3H_{10} \div {}_4H_{10} = {}_{12}C_{10} \div {}_{13}C_{10} = \frac{12 \cdot 11 / 2}{13 \cdot 12 \cdot 11 / 6} = \frac{3}{13}.$$ The ${}_3H_{10}$ is simply if you can't choose the blue suit but it looks like you misinterpret the $r$ to be $9$.

Sadly if the question doesn't mention something "one by one" then you should take all the $10$ cards by once $-$ and thus you should choose between ${}_nP_r$ and ${}_nC_r$. Now, to "find the probability that at least one blue card was chosen", you just don't need to care about the order of the cards.

To illustrate that order doesn't matter, let's just shrink the deck size to $12$ (so each suit has $3$) and no. of cards to pick to $3$.

• If you care the order (i.e. to see Yellow-Green-Red and Green-Yellow-Red as $2$ different cases), you have $12 \cdot (12 - 1) \cdot (12 - 2) = 1320$ choices to pick $3$ cards, and you have $(12 - 2) \cdot (12 - 2 - 1) \cdot (12 - 2 - 2) = 720$ choices to not to pick blue cards. The required probability then becomes $1 - 720/1320 = 5/11.$
• Meanwhile if you don't care the order (i.e. to see Yellow-Green-Red and Green-Yellow-Red as the same case), you have to divide the choices you just calculated by $3! -$ the same 3 cards has $3!$ orders to be chosen. Then you have $12 \cdot (12 - 1) \cdot (12 - 2) / 3! = 220$ choices to pick $2$ cards and $(12 - 2) \cdot (12 - 2 - 1) \cdot (12 - 2 - 2) / 3! = 120$ choices to not to pick blue cards. The required probability then becomes $1 - 120/220 = 5/11.$

No matter you consider the order to pick the cards or not, you get the same value of the probability. Thus don't care the order and use ${}_nC_r$. What do you think?