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We have a deck of 52 cards we split it in to two decks. First deck contains 20 red and 14 black cards, second deck contains 6 red and 12 black card, from each deck we pull a card. From these two cards we pick one of them.

What is the probability of getting a red card in the end?

Let $P(A_{1})$ be the probability of getting a red card out of both decks
$$ P(A_{1})= \frac{\binom{34}{20}+\binom{18}{6}-\binom{34}{20}\binom{18}{6}}{\binom{52}{1}} $$ $\binom{18}{6}$ is the chance of getting a red card out of the second deck, $\binom{18}{6}*\binom{34}{20}$ is the chance of getting intercepted, which we don't want and finally $\binom{34}{20}$ is the chance of getting a red card out the first deck.

After that We have to calculate the probability of getting a red card from the first deck and getting a black from the second, Let P($A_{2}$) be the probability of that: $$P(A_{2})=\frac{\binom{34}{20}*\binom{18}{12}}{52}*0.5$$ Similar situation here, $\binom{34}{20}$ give the probability of getting a red card from one deck, from the other we get a black card and we *0.5 for obvious reasons.

Analogically we do the same for $A_{3}$ : $$P(A_{3})=0.5*\frac{\binom{34}{14}*\binom{18}{6}}{\binom{52}{1}}$$

So the final answer is $P(A_{1})+P(A_{2})+P(A_{3})$ Is that correct ?

One more thing we have to find the probability of pulling a red card which came from the first deck: $$P(A_{2})=\frac{\binom{34}{20}*1}{52}*0.5$$

Are my calculations correct and what do I have to do if they aren't

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    Seems simpler to write, for example, that the probability of getting red from each pile is $\frac {20}{34}\times \frac 6{18}$. I don't even think your expression is positive. – 2017-02-05
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    The notation $\binom{18}{6}$ (which equals 18564) means that you want to choose 6 items out of 18 available. The probability of choosing one red card from the second deck is $6/18$. – 2017-02-05
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    Please clarify: Your task is to split a deck of cards as indicated, choose a card from each sub-deck, then randomly choose one of the two cards, and finally determine the probability of that card being red. Am I correct? – 2017-02-05
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    yes, Tim you understand correctly..about the binomal, picking a number of cards from a deck, can't it be expressed with combinations? – 2017-02-05
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    picking one card from 52 isn't it equal to $\binom{52}{1}$ – 2017-02-05
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    The expression $\binom{52}{1}$ calculates the **number** of one-card combinations you have out of 52 cards. That isn't a probability. You can use combinations to find probabilities. For example, start with a standard 52-card deck. Then the probability of selecting one red card would be $\binom{26}{1}/\binom{52}{1}$. – 2017-02-05

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Assuming I am correctly interpreting the question asked...

The probability of choosing a red card from the first deck: $$P(\text{red card from first deck})=\frac{20}{34}$$

The probability of choosing a red card from the second deck: $$P(\text{red card from second deck})=\frac{6}{18}$$

Randomly selecting one of the cards means a $1/2$ probability for either card, so the probability of selecting a red card: $$P(\text{red card})=0.5\cdot\frac{20}{34}+0.5\cdot\frac{6}{18}\approx0.4608$$

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    Are you sure because I think I have to solve the problem with full probability, and what do you think about the second question? What is the probability that we pick a red card and.. it came from the first deck ? – 2017-02-05
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    For your second question, use conditional probability. What do you mean by "full probability"? – 2017-02-05
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    Full probability- a way of solving the problem, considering all possible outcomes. I know about the second one but is my answer correct ? – 2017-02-05
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    I don't see where you asked your second question nor do I see your answer... – 2017-02-05