If A, B, C are three events such that `P(B)=(4)/(5), P(A nn B nn C^(c ))=(1)/(4) and P(A^(c )nnBnnC^(c ))=(1)/(3)`, then `P(BnnC)` is equal to

Options

A. `(11)/(60)`
B. `(1)/(5)`
C. `(13)/(60)`
D. `(1)/(4)`

Answer

To solve the problem, we need to find \( P(B \cap C) \) using the given probabilities. Let's break down the information provided and derive the solution step by step.

### Given:
1. \( P(B) = \frac{4}{5} \)
2. \( P(A \cap B \cap C^c) = \frac{1}{4} \)
3. \( P(A^c \cap B \cap C^c) = \frac{1}{3} \)

### Step 1: Understand the events
We have three events A, B, and C. The notation \( C^c \) represents the complement of event C (the event that C does not occur).

### Step 2: Use the probabilities
We can express the total probability of event B as follows:
\[ P(B) = P(A \cap B \cap C) + P(A \cap B \cap C^c) + P(A^c \cap B \cap C) + P(A^c \cap B \cap C^c) \]

### Step 3: Substitute known values
From the information given:
- We know \( P(A \cap B \cap C^c) = \frac{1}{4} \)
- We know \( P(A^c \cap B \cap C^c) = \frac{1}{3} \)

Let:
- \( x = P(A \cap B \cap C) \)
- \( y = P(A^c \cap B \cap C) \)

Now we can rewrite the equation for \( P(B) \):
\[ P(B) = x + \frac{1}{4} + y + \frac{1}{3} \]

### Step 4: Find a common denominator
To combine \( \frac{1}{4} \) and \( \frac{1}{3} \), we find a common denominator:
- The least common multiple of 4 and 3 is 12.
- Thus, \( \frac{1}{4} = \frac{3}{12} \) and \( \frac{1}{3} = \frac{4}{12} \).

So,
\[ \frac{1}{4} + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \]

### Step 5: Substitute back into the equation
Now we have:
\[ P(B) = x + y + \frac{7}{12} \]
Given \( P(B) = \frac{4}{5} \), we can equate:
\[ \frac{4}{5} = x + y + \frac{7}{12} \]

### Step 6: Convert \( \frac{4}{5} \) to a fraction with a denominator of 60
To solve for \( x + y \), we convert \( \frac{4}{5} \) to have a denominator of 60:
\[ \frac{4}{5} = \frac{48}{60} \]
And \( \frac{7}{12} \) to have a denominator of 60:
\[ \frac{7}{12} = \frac{35}{60} \]

### Step 7: Substitute and solve for \( x + y \)
Now we have:
\[ \frac{48}{60} = x + y + \frac{35}{60} \]
Subtract \( \frac{35}{60} \) from both sides:
\[ x + y = \frac{48}{60} - \frac{35}{60} = \frac{13}{60} \]

### Step 8: Find \( P(B \cap C) \)
Now we can find \( P(B \cap C) \):
\[ P(B) = P(B \cap C) + P(B \cap C^c) \]
Where \( P(B \cap C^c) = P(A \cap B \cap C^c) + P(A^c \cap B \cap C^c) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \).

Thus,
\[ P(B \cap C) = P(B) - P(B \cap C^c) = \frac{4}{5} - \frac{7}{12} \]
Convert \( \frac{4}{5} \) to have a denominator of 60:
\[ \frac{4}{5} = \frac{48}{60} \]
Convert \( \frac{7}{12} \) to have a denominator of 60:
\[ \frac{7}{12} = \frac{35}{60} \]
Now,
\[ P(B \cap C) = \frac{48}{60} - \frac{35}{60} = \frac{13}{60} \]

### Final Answer:
\[ P(B \cap C) = \frac{13}{60} \]