If $p = \frac{1}{8}; n = 640; q = \frac{7}{8}$, then variance Binomial Distribution
Step 1: Recall the formula for the variance of a binomial distribution.
A binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success) has variance given by: \[ \text{Variance} = n p q, \] where \( q = 1 - p \) is the probability of failure.
Step 2: Identify the given values.
\( p = \frac{1}{8} \),
\( q = \frac{7}{8} \) (which satisfies \( q = 1 - p \), since \( 1 - \frac{1}{8} = \frac{7}{8} \)),
\( n = 640 \).
Step 3: Compute the variance.
\[ \text{Variance} = n p q = 640 \cdot \frac{1}{8} \cdot \frac{7}{8}. \] First, calculate: \[ \frac{1}{8} \cdot \frac{7}{8} = \frac{7}{64}, \] \[ \text{Variance} = 640 \cdot \frac{7}{64} = \frac{640 \cdot 7}{64} = \frac{4480}{64} = 70. \]
Step 4: Evaluate the options.
(1) 0.07: Incorrect, as the variance is 70, not 0.07. Incorrect.
(2) 0.7: Incorrect, as the variance is 70, not 0.7. Incorrect.
(3) 7.0: Incorrect, as the variance is 70, not 7.0. Incorrect.
(4) 70.0: Correct, as the variance is 70. Correct.
Step 5: Select the correct answer.
The variance of the binomial distribution is 70.0, matching option (4).
P and Q play chess frequently against each other. Of these matches, P has won 80% of the matches, drawn 15% of the matches, and lost 5% of the matches.
If they play 3 more matches, what is the probability of P winning exactly 2 of these 3 matches?
If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)