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?

To find \( P(B) \) given that events \( A \) and \( B \) are independent, we use the formula for the union of two independent events: \[ P(A \cup B) = P(A) + P(B) - P(A)P(B) \] Substituting the known values, we have: \[ 0.89 = 0.21 + P(B) - 0.21 P(B) \] Let \( P(B) = x \). Rearranging the equation gives: \[ 0.89 = 0.21 + x - 0.21x \] Which simplifies to: \[ 0.89 - 0.21 = x(1 - 0.21) \implies 0.68 = 0.79x \implies x = \frac{0.68}{0.79} \approx 0.8608 \] Thus, \( P(B) \approx 0.861 \). --- To find the probability of drawing a red ball first and a yellow ball second from the box without replacement, we can use the multiplication rule for dependent events: 1. The probability of drawing a red ball first is \( \frac{2}{6} \). 2. After removing one red ball, there are now 5 balls left, of which 3 are yellow. Thus, the probability of drawing a yellow ball next is \( \frac{3}{5} \). Now multiply these probabilities: \[ P(\text{Red first, Yellow second}) = \frac{2}{6} \times \frac{3}{5} = \frac{2 \times 3}{6 \times 5} = \frac{6}{30} = \frac{1}{5} = 0.2 \] Hence, the probability that the first ball will be red and the second one yellow is \( 0.2 \) or 20%.