The problem states that a bag contains green, red, and blue balls. Anna picks a ball at random and puts it back. We are given a table with the probabilities of picking a green ball (0.55) and a red ball (0.25). The probability of picking a blue ball is represented by $x$. (a) We need to find the value of $x$. (b) We are told there are 20 red balls in the bag and we need to find the number of green balls in the bag. (c) We are asked to complete a probability tree diagram, but the tree diagram is not provided in the image, so it cannot be completed.

1. Problem Description

The problem states that a bag contains green, red, and blue balls. Anna picks a ball at random and puts it back. We are given a table with the probabilities of picking a green ball (0.55) and a red ball (0.25). The probability of picking a blue ball is represented by $x$.

(a) We need to find the value of $x$.

(b) We are told there are 20 red balls in the bag and we need to find the number of green balls in the bag.

(c) We are asked to complete a probability tree diagram, but the tree diagram is not provided in the image, so it cannot be completed.

2. Solution Steps

(a) To find $x$, we use the fact that the sum of probabilities of all possible outcomes is

1. $P(green) + P(red) + P(blue) = 1$

$0.55 + 0.25 + x = 1$

$0.80 + x = 1$

$x = 1 - 0.80$

$x = 0.20$

(b) Let $N_{green}$, $N_{red}$, and $N_{blue}$ be the number of green, red, and blue balls respectively. Let $N_{total}$ be the total number of balls. We are given $N_{red} = 20$.

The probability of picking a red ball is $P(red) = \frac{N_{red}}{N_{total}}$. We know $P(red) = 0.25$, so $0.25 = \frac{20}{N_{total}}$.

$N_{total} = \frac{20}{0.25} = 80$

The probability of picking a green ball is $P(green) = \frac{N_{green}}{N_{total}}$. We know $P(green) = 0.55$, so $0.55 = \frac{N_{green}}{80}$.

$N_{green} = 0.55 \times 80 = 44$

(c) Cannot be completed as the tree diagram is not available.

3. Final Answer

(a) $x = 0.20$

(b) The number of green balls is

4

4. (c) Cannot be completed.