There are only crange marbles and blue marbles in a bag. There are 6 arange marbles and 4 blue marbles. Lidija fakes a marble from the bag at random. She puts the marble back in the bag She then takes at random another marble from the bag. What is the probability of not choosing a blue marble from the bag? TIP: You need to construct a tree diagram and use it to calculate the probability for this event. INSTRUCTION: Your answer must be a fraction in its lowest terms. Answer: The probability of not choosing a blue marble from the bag is

To solve this, we can think of two draws from the bag and the total number of marbles present. There are 6 orange and 4 blue marbles, making a total of 10 marbles. 1. The probability of drawing an orange marble (not blue) on the first draw is \( \frac{6}{10} = \frac{3}{5} \). 2. The probability of drawing a blue marble on the first draw is \( \frac{4}{10} = \frac{2}{5} \). 3. Since Lidija puts the marble back, the probabilities for the second draw remain unchanged. Now, we calculate the probabilities of the different branches: - **Branch 1:** First draw orange, second draw orange: \(\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}\) - **Branch 2:** First draw orange, second draw blue: \(\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}\) - **Branch 3:** First draw blue, second draw orange: \(\frac{2}{5} \times \frac{3}{5} = \frac{6}{25}\) - **Branch 4:** First draw blue, second draw blue: \(\frac{2}{5} \times \frac{2}{5} = \frac{4}{25}\) Now, if we are looking for the probability of **not** choosing a blue marble in both draws, we only consider **Branch 1** (orange, orange) and **Branch 2** (orange, blue) and **Branch 3** (blue, orange), because those involve at least one orange. But since we want the result of not picking a blue marble in any draw, we only focus on **Branch 1**. Therefore, the required probability is: \[ P(\text{not blue on both draws}) = \frac{9}{25} \] The probability of not choosing a blue marble from the bag is \(\frac{9}{25}\).