In a pack of 52 cards , a card is missing . If 2 cards are drawn randomly and found to be of spades . Then probability thatmissing card is not of spade

Answer

To solve the problem, we need to find the probability that the missing card is not a spade given that two cards drawn are spades. We can use Bayes' theorem to find this probability.

### Step 1: Define the events
Let:
- \( A \): The event that the missing card is a spade.
- \( B \): The event that two cards drawn are spades.

We need to find \( P(A^c | B) \), where \( A^c \) is the event that the missing card is not a spade.

### Step 2: Use Bayes' theorem
According to Bayes' theorem:
\[ P(A^c | B) = \frac{P(B | A^c) \cdot P(A^c)}{P(B)} \]

### Step 3: Calculate \( P(A^c) \)
In a standard deck of 52 cards, there are 13 spades. Therefore, the probability that the missing card is not a spade is:
\[ P(A^c) = \frac{39}{52} = \frac{3}{4} \]

### Step 4: Calculate \( P(B | A^c) \)
If the missing card is not a spade, there are still 13 spades in the remaining 51 cards. The probability of drawing 2 spades from these 51 cards can be calculated using combinations:
\[ P(B | A^c) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{\frac{13 \times 12}{2}}{\frac{51 \times 50}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550} = \frac{26}{425} \]

### Step 5: Calculate \( P(B | A) \)
If the missing card is a spade, there are only 12 spades left in the remaining 51 cards. Thus:
\[ P(B | A) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{\frac{12 \times 11}{2}}{\frac{51 \times 50}{2}} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550} = \frac{22}{425} \]

### Step 6: Calculate \( P(B) \)
Using the law of total probability:
\[ P(B) = P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c) \]
Where \( P(A) = \frac{13}{52} = \frac{1}{4} \):
\[ P(B) = \left(\frac{22}{425} \cdot \frac{1}{4}\right) + \left(\frac{26}{425} \cdot \frac{3}{4}\right) \]
\[ P(B) = \frac{22}{1700} + \frac{78}{1700} = \frac{100}{1700} = \frac{1}{17} \]

### Step 7: Substitute values into Bayes' theorem
Now we can substitute back into Bayes' theorem:
\[ P(A^c | B) = \frac{P(B | A^c) \cdot P(A^c)}{P(B)} = \frac{\left(\frac{26}{425}\right) \cdot \left(\frac{3}{4}\right)}{\frac{1}{17}} \]
\[ = \frac{\frac{78}{1700}}{\frac{1}{17}} = \frac{78}{1700} \cdot 17 = \frac{78}{100} = 0.78 \]

### Final Answer
Thus, the probability that the missing card is not a spade is:
\[ \boxed{0.78} \]