Probability Calculation for Card Drawing

We need to find the probability of drawing two cards from a standard 52-card deck such that one card is red and the other is black.

Method: Sequential Probability

Consider drawing the cards one after the other without replacement.

1. Possible Scenarios: There are two ways this can happen:
   • Drawing a red card first, then a black card.
   • Drawing a black card first, then a red card.
2. Scenario 1: Red then Black
   • The probability of the first card being red is $\frac{26}{52}$ (since there are 26 red cards out of 52 total).
   • After drawing one red card, there are 51 cards left. The probability of the second card being black is $\frac{26}{51}$ (since there are still 26 black cards).
   • The probability of this sequence (Red then Black) is: $P(\text{Red then Black}) = \frac{26}{52} \times \frac{26}{51}$
3. Scenario 2: Black then Red
   • The probability of the first card being black is $\frac{26}{52}$.
   • After drawing one black card, there are 51 cards left. The probability of the second card being red is $\frac{26}{51}$.
   • The probability of this sequence (Black then Red) is: $P(\text{Black then Red}) = \frac{26}{52} \times \frac{26}{51}$
4. Total Probability: The total probability of drawing one red and one black card is the sum of the probabilities of these two mutually exclusive scenarios: $P(\text{One Red, One Black}) = P(\text{Red then Black}) + P(\text{Black then Red})$ $P(\text{One Red, One Black}) = \left( \frac{26}{52} \times \frac{26}{51} \right) + \left( \frac{26}{52} \times \frac{26}{51} \right)$ $P(\text{One Red, One Black}) = 2 \times \left( \frac{26}{52} \times \frac{26}{51} \right)$ $P(\text{One Red, One Black}) = 2 \times \left( \frac{1}{2} \times \frac{26}{51} \right)$ $P(\text{One Red, One Black}) = \frac{26}{51}$

Combinations Method (Alternative)

Alternatively, using combinations:

• Total ways to choose 2 cards from 52: $C(52, 2) = \frac{52 \times 51}{2 \times 1} = 1326$.
• Ways to choose 1 red card from 26: $C(26, 1) = 26$.
• Ways to choose 1 black card from 26: $C(26, 1) = 26$.
• Ways to choose 1 red AND 1 black card: $C(26, 1) \times C(26, 1) = 26 \times 26 = 676$.
• Probability = $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{676}{1326}$.
• Simplifying the fraction: $\frac{676}{1326} = \frac{338}{663} = \frac{26 \times 13}{51 \times 13} = \frac{26}{51}$.

Conclusion

The probability of drawing one red card and one black card is $\frac{26}{51}$.