To find \( P(A \cup B) \), we can use the formula of probability for the union of two events: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
First, we need to find \( P(A \cap B) \), which is the intersection of events \( A \) and \( B \).
We use the conditional probability formula: \( P(A \cap B) = P(A | B) \cdot P(B) \).
Given \( P(A | B) = 0.6 \) and \( P(B) = 0.8 \), we compute \( P(A \cap B) \) as follows:
\[ P(A \cap B) = 0.6 \times 0.8 = 0.48 \]
Now we can find \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
\[ P(A \cup B) = 0.4 + 0.8 - 0.48 = 0.72 \]
Therefore, the correct answer is 0.72.
The formula for the union of two events is:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).
The conditional probability \( P(A \mid B) \) is related to \( P(A \cap B) \) as:
\( P(A \cap B) = P(A \mid B) \cdot P(B) \).
Substitute the given values:
\( P(A \cap B) = (0.6)(0.8) = 0.48 \).
Now calculate \( P(A \cup B) \):
\( P(A \cup B) = 0.4 + 0.8 - 0.48 = 0.72 \).
Thus, the probability \( P(A \cup B) \) is 0.72.
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: \( P(3 < x \leq 6) \)
If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :