teach IGCSE student how to solve the questions, the workprocess---Question Number: 5 Problem Description: A box contains 3 blue pens and 5 red pens. (a) Mia picks a pen from the box at random. Find the probability that she picks a red pen. [1] (b) Mia puts the pen back into the box. She then picks a pen at random and replaces it. She then picks a second pen at random. (i) Complete the tree diagram. [2] (ii) Find the probability that Mia picks two pens that have the same colour. [3] Chart Description: Type: Tree diagram. Main Elements: - Two main stages labelled "First Pen" and "Second Pen". - From the "First Pen" node, there are two branches. - The upper branch is labelled "Blue". A dotted line is next to this branch, indicating a missing probability value. - The lower branch is labelled "Red". A dotted line is next to this branch, indicating a missing probability value. - From the "Blue" node in the "Second Pen" column (reached via the upper branch from "First Pen"), there are two branches. - The upper branch is labelled "Blue". A dotted line is next to this branch, indicating a missing probability value. - The lower branch is labelled "Red". A dotted line is next to this branch, indicating a missing probability value. - From the "Red" node in the "Second Pen" column (reached via the lower branch from "First Pen"), there are two branches. - The upper branch is labelled "Blue". A dotted line is next to this branch, indicating a missing probability value. - The lower branch is labelled "Red". A dotted line is next to this branch, indicating a missing probability value. - The diagram is structured to show the possible outcomes of picking a pen in the first stage and then picking a pen in the second stage, with replacement. The dotted lines represent the probabilities of each branch occurring. (c) Mia now picks 3 of the 8 pens in the box at random without replacement. Find the probability that she picks 2 blue pens and 1 red pen. [3]

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Let's solve this probability problem step by step. We have a box with 3 blue pens and 5 red pens. For part a, we need to find the probability of picking a red pen at random. First, we count the total number of pens: 3 plus 5 equals 8 pens. Next, we identify that there are 5 red pens. Finally, we calculate the probability as the number of red pens divided by the total number of pens, which gives us 5 over 8. Now let's work on part b. Mia puts the pen back and picks two pens with replacement. This means after picking the first pen, she returns it to the box before picking the second pen. Since the pen is replaced, the probabilities remain the same for each pick. The probability of picking blue is 3 over 8, and the probability of picking red is 5 over 8. Because of replacement, these same probabilities apply to both the first and second picks, regardless of what was picked first. Now for part b ii, we need to find the probability that Mia picks two pens of the same color. Same color can happen in two ways: blue then blue, or red then red. For blue then blue, we multiply 3 over 8 times 3 over 8, which equals 9 over 64. For red then red, we multiply 5 over 8 times 5 over 8, which equals 25 over 64. Since these are mutually exclusive events, we add the probabilities: 9 over 64 plus 25 over 64 equals 34 over 64, which simplifies to 17 over 32. For part c, Mia picks 3 pens without replacement. This means each pick changes the number of remaining pens. We want 2 blue and 1 red pen. There are three possible sequences: blue blue red, blue red blue, and red blue blue. For sequence blue blue red: first pick is 3 over 8, second pick is 2 over 7 since one blue was removed, third pick is 5 over 6 since two pens were removed. This gives 30 over 336. Similarly, the other two sequences also give 30 over 336 each. Adding all three: 90 over 336, which simplifies to 15 over 56. Let's summarize our complete solutions. For part a, the probability of picking a red pen is 5 over 8. For part b i, we completed the tree diagram with probabilities 3 over 8 for blue and 5 over 8 for red at each stage, since replacement keeps the probabilities constant. For part b ii, the probability of picking two pens of the same color is 17 over 32, found by adding the blue-blue and red-red outcomes. For part c, the probability of picking 2 blue and 1 red pen without replacement is 15 over 56, calculated by summing three possible sequences. The key concept is understanding how replacement versus non-replacement affects probability calculations.