# A probability problem - probability of one card being red and the other one being black.

Consider a deck of 50 playing cards (2 cards missing). What is the probability that one of them is red and the other one is black? I've got two solutions which one is correct ?

Let $R$ represent red card and $B$ black.

$\therefore$ sample space = $\{RR, BB, RB, BR\}$

favorable outcomes = $\{RB, BR\}$

Probability = ${2 \over 4}=0.5$

But $BR$ is same as $RB$,

$\therefore$ sample space = $\{RR, BB, RB\}$

favorable outcomes = $\{RB\}$

$\therefore$ probability should be $1 \over 3$.

Which one is correct.

I think it should be $1 \over 3$. Because $BR$ should be taken same as $RB$ but when we talk about two coins being flipped and the probability of both of them showing head is $1 \over 4$, right? Here sample space becomes $\{HH, TT, TH, HT\}$. So why both $HT$ and $TH$ is taken?

My question is while calculating probability, when do we include or exclude same combinations? Like $RB, BR$ or $HT, TH$...

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I'm assuming you pick one card, look at it, replace it, shuffle, pick another card. Then the probability that the two cards have different colours is $\frac{1}{2}$.

If $N$ is the number of elements in the sample space, then $\frac{1}{N}$ is the probability of an outcome only if all outcomes are equally likely.

In your second set {RR,BB,RB}, RB is twice as likely to occur as RR (and RR is as likely to occur as RB).

Look at it another way. You could call your sample set {both same colour, one of each colour}. The event {both same colour} is made up of outcomes RR and BB. The event {one of each colour} is made up of outcomes RB and BR.

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One considers a deck of N red cards and N black cards. One draws two cards from this deck, without replacement. One is asking for the probability p(N) that the second card drawn from the deck has a different color than the first card drawn.

After the first card is drawn, there remains 2N-1 cards in the deck and amongst those, N cards have a color different than the color of the first card drawn. Hence p(N) = N/(2N-1).

In particular, p(26) = 26/51, which is just below 51%.

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