To solve the problem of finding \( P\left( B | \left( A \cup \overline{B} \right) \right) \), we will use some basic probability principles and formulas.
Therefore, the probability \( P(B | A \cup \overline{B}) \) is \(\frac{1}{4}\), which matches the correct answer option.
Step 1: Find \( P(A \cap B) \)
\[ P(A \cap \overline{B}) = P(A) - P(A \cap B) \]
\[ 0.5 = 0.7 - P(A \cap B) \]
\[ P(A \cap B) = 0.2 \]
Step 2: Calculate \( P(A \cup \overline{B}) \)
Using probability rules:
\[ P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B}) \]
\[ = 0.7 + (1 - 0.4) - 0.5 \]
\[ = 0.7 + 0.6 - 0.5 = 0.8 \]
Step 3: Find \( P(B \cap (A \cup \overline{B})) \)
\[ P(B \cap (A \cup \overline{B})) = P((B \cap A) \cup (B \cap \overline{B})) \]
\[ = P(A \cap B) + P(\emptyset) = 0.2 + 0 = 0.2 \]
Step 4: Compute conditional probability
\[ P(B | A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{0.2}{0.8} = \frac{1}{4} \]
A board has 16 squares as shown in the figure. Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)