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
\[ P(B | A \cup \overline{B}) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{0.2}{0.8} = \frac{1}{4} \]
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :
If the probability distribution is given by:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| P(x) | 0 | k | 2k | 2k | 3k | k² | 2k² | 7k² + k |
Then find: \( P(3 < x \leq 6) \)
Find the area of the region \[ R = \{(x, y) : xy \le 27,\; 1 \le y \le x^2 \}. \]
An object of uniform density rolls up the curved path with the initial velocity $v_o$ as shown in the figure. If the maximum height attained by an object is $\frac{7v_o^2}{10 g}$ ($g=$ acceleration due to gravity), the object is a _______

A body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth ($R_e$). The increase in potential energy will be ____ ($g$ is acceleration due to gravity at the surface of earth)