While shuffling a pack of 52 cards, 2 cards are accidently dropped. The probability that missing cards are of different colours is

Options

A. `26/51`
B. `25/51`
C. `1/2`
D. `25/52`

Answer

To solve the problem of finding the probability that the two missing cards from a pack of 52 cards are of different colors, we can follow these steps:

### Step 1: Understand the Composition of the Deck
A standard deck of 52 cards consists of:
- 26 red cards (hearts and diamonds)
- 26 black cards (clubs and spades)

### Step 2: Define the Event
We need to find the probability that the two dropped cards are of different colors. This can happen in two scenarios:
1. The first card is red and the second card is black.
2. The first card is black and the second card is red.

### Step 3: Calculate the Probability for Each Scenario
**Scenario 1: First card is red, second card is black**
- The probability of drawing a red card first:
\[ P(\text{Red first}) = \frac{26}{52} \]
- After drawing one red card, there are now 51 cards left, with 26 of them being black. The probability of drawing a black card second:
\[ P(\text{Black second | Red first}) = \frac{26}{51} \]
- Therefore, the combined probability for this scenario is:
\[ P(\text{Red first and Black second}) = P(\text{Red first}) \times P(\text{Black second | Red first}) = \frac{26}{52} \times \frac{26}{51} \]

**Scenario 2: First card is black, second card is red**
- The probability of drawing a black card first:
\[ P(\text{Black first}) = \frac{26}{52} \]
- After drawing one black card, there are still 26 red cards left among the 51 remaining cards. The probability of drawing a red card second:
\[ P(\text{Red second | Black first}) = \frac{26}{51} \]
- Therefore, the combined probability for this scenario is:
\[ P(\text{Black first and Red second}) = P(\text{Black first}) \times P(\text{Red second | Black first}) = \frac{26}{52} \times \frac{26}{51} \]

### Step 4: Combine the Probabilities
Now, we can add the probabilities of both scenarios to find the total probability of the two cards being of different colors:
\[ P(\text{Different colors}) = P(\text{Red first and Black second}) + P(\text{Black first and Red second}) \]
\[ = \left(\frac{26}{52} \times \frac{26}{51}\right) + \left(\frac{26}{52} \times \frac{26}{51}\right) \]
\[ = 2 \times \left(\frac{26}{52} \times \frac{26}{51}\right) \]
\[ = 2 \times \frac{26 \times 26}{52 \times 51} \]

### Step 5: Simplify the Expression
Now we simplify:
\[ = 2 \times \frac{676}{2652} \]
\[ = \frac{1352}{2652} \]
Now, we can simplify \(\frac{1352}{2652}\) by dividing both the numerator and the denominator by 52:
\[ = \frac{26}{51} \]

### Final Answer
Thus, the probability that the two missing cards are of different colors is:
\[ \boxed{\frac{26}{51}} \]