To solve the problem, we need to find the probability \(P(A|B)\), where event \(A\) is that the first ball is black, and event \(B\) is that the second ball is black. The probability formula for conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Step 1: Find \(P(A \cap B)\).
The probability that the first ball is black and the second ball is also black can be calculated by considering the following:
Therefore:
\[ P(A \cap B) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \]
Step 2: Find \(P(B)\).
\(P(B)\) is the probability that the second ball is black regardless of the color of the first ball. Consider the two scenarios:
Thus:
\[ P(B) = \left(\frac{6}{10} \times \frac{5}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{30}{90} + \frac{24}{90} = \frac{54}{90} = \frac{3}{5} \]
Step 3: Calculate \(P(A|B)\).
Using the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3} \times \frac{5}{3} = \frac{5}{9} \]
With gcd(5, 9) = 1, the fraction \(\frac{m}{n}\) is in its simplest form with \(m = 5\) and \(n = 9\). Hence, \(m + n\) equals \(14\).
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).