Events A and B are such that \( P(A)=0,21 \) and \( P(A \cup B)=0,89 \). Let's assume that events \( A \) and \( B \) are independent, determine \( P(B) \). Question 5 A box contains 6 balls of the same size of which 2 are red, 3 are yellow and 1 is white. If 2 balls are selected at random from the box without replacement, what is the probability that the first ball will be red and second one yellow?

Sure, let's tackle each question step by step. --- ## **Question 1** **Given:** - \( P(A) = 0.21 \) - \( P(A \cup B) = 0.89 \) - Events \( A \) and \( B \) are **independent**. **Find:** \( P(B) \) ### **Solution:** When two events \( A \) and \( B \) are independent, the probability of their intersection is the product of their probabilities: \[ P(A \cap B) = P(A) \cdot P(B) \] The probability of the union of two events is given by: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the expression for \( P(A \cap B) \) into the union formula: \[ P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B) \] We know \( P(A) = 0.21 \) and \( P(A \cup B) = 0.89 \). Plug these values into the equation: \[ 0.89 = 0.21 + P(B) - (0.21 \cdot P(B)) \] Simplify the equation: \[ 0.89 = 0.21 + P(B) - 0.21P(B) \] \[ 0.89 - 0.21 = P(B)(1 - 0.21) \] \[ 0.68 = 0.79P(B) \] Now, solve for \( P(B) \): \[ P(B) = \frac{0.68}{0.79} \] \[ P(B) \approx 0.8608 \] **Answer:** \( P(B) \approx 0.8608 \) or 86.08% --- ## **Question 5** **Given:** - Total balls in the box: 6 - Red balls: 2 - Yellow balls: 3 - White balls: 1 - Two balls are selected at random **without replacement**. - Find the probability that: 1. The **first** ball is **red**. 2. The **second** ball is **yellow**. **Find:** \( P(\text{First Red and Second Yellow}) \) ### **Solution:** This is a compound probability problem without replacement. We'll use the multiplication rule for dependent events. 1. **Probability that the first ball is red:** \[ P(\text{First Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{2}{6} = \frac{1}{3} \] 2. **After drawing a red ball, update the counts:** - Remaining balls: \( 6 - 1 = 5 \) - Remaining red balls: \( 2 - 1 = 1 \) - Yellow balls remain unchanged: 3 3. **Probability that the second ball is yellow given the first was red:** \[ P(\text{Second Yellow | First Red}) = \frac{\text{Number of yellow balls}}{\text{Remaining total balls}} = \frac{3}{5} \] 4. **Combined probability using the multiplication rule:** \[ P(\text{First Red and Second Yellow}) = P(\text{First Red}) \times P(\text{Second Yellow | First Red}) \] \[ = \frac{1}{3} \times \frac{3}{5} \] \[ = \frac{1 \times 3}{3 \times 5} \] \[ = \frac{3}{15} \] \[ = \frac{1}{5} \] \[ = 0.2 \] \[ = 20\% \] **Answer:** The probability is \( \frac{1}{5} \) or 20%. --- Feel free to ask if you have any further questions or need additional explanations!