Two cards are drawn at random and one by one without replacement from a well shuffled

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Text Solution

Solution :  Total given number of cards =52<br> Here, you have to find probability of choosing a ref card and a black card.<br> P( one red and one black) =p( first red and second black )+p( first black & second red)<br> `=frac{26}{52} times frac{26}{51}+frac{26}{52} times frac{26}{51}`<br> (without replacement)<br> `=frac{13}{51}+frac{13}{51}`<br> `=frac{26} {51}`<br>

Your answer would be correct if the question was asking the probability of three draws with replacement, with the first card being a King, the second card being a Queen and the third card being a Jack - in other words, if the order in which $K, Q$ and $J$ should appear were fixed. But for the given question, they can appear in any order. So you need to multiple your answer by $3!$. Alternatively, you can think of it as follows.

First card can be any of the $12$ cards of ranks $K, Q$ and $J$. Depending on the rank of the first card, the next card has to be one of the $8$ of other two ranks and the last card has to be from the $4$ cards of the last rank.

So the probability is $~ \displaystyle \frac{12}{52} \cdot \frac{8}{52} \cdot \frac{4}{52} = \frac{6}{13^3}$

Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?

\[P\left( \text{ one red and one black } \right) = P\left(\text{  first red and second black } \right) + P\left( \text{ first black and second red }  \right)\]\[ = \frac{26}{52} \times \frac{26}{51} + \frac{26}{52} \times \frac{26}{51} \left[ \text{ Without replacement }  \right]\]\[ = \frac{13}{51} + \frac{13}{51}\]

\[ = \frac{26}{51}\]

Concept: Probability Examples and Solutions

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