Question: Write all your answers as FRACTIONS in SIMPLEST FORM. When drawing TWO M&Ms from a jar WITH…

Write all your answers as FRACTIONS in SIMPLEST FORM.

When drawing TWO M&Ms from a jar WITH REPLACEMENT, find the PROBABILITY of getting:

1. A red and a red. \( P(R \cap R) = \)
2. A red and a yellow. \( P(R \cap Y) = \)
3. A yellow and a yellow. \( P(Y \cap Y) = \)
4. A blue and a red. \( P(B \cap R) = \)

When drawing TWO M&Ms from a jar WITHOUT REPLACEMENT, find the PROBABILITY of getting:

5. A red and a red. \( P(R \cap R) = \)
6. A red and a yellow. \( P(R \cap Y) = \)
7. A yellow and a yellow. \( P(Y \cap Y) = \)
8. A blue and a red. \( P(B \cap R) = \)

8 Yellow 4 Red 9 Blue

Solution

To solve the probability problems, we need to determine the probabilities of drawing different combinations of M&Ms with and without replacement. Given: - 8 Yellow M&Ms - 4 Red M&Ms - 9 Blue M&Ms Total M&Ms \( = 8 + 4 + 9 = 21 \) ### With Replacement 1. A red and a red Probability of drawing a red first \( = \frac{4}{21} \) Probability of drawing a red second \( = \frac{4}{21} \) Total probability: \[ P(R \cap R) = \frac{4}{21} \times \frac{4}{21} = \frac{16}{441} \] 2. A red and a yellow Probability of drawing a red first \( = \frac{4}{21} \) Probability of drawing a yellow second \( = \frac{8}{21} \) Total probability: \[ P(R \cap Y) = \frac{4}{21} \times \frac{8}{21} = \frac{32}{441} \] 3. A yellow and a yellow Probability of drawing a yellow first \( = \frac{8}{21} \) Probability of drawing a yellow second \( = \frac{8}{21} \) Total probability: \[ P(Y \cap Y) = \frac{8}{21} \times \frac{8}{21} = \frac{64}{441} \] 4. A blue and a red Probability of drawing a blue first \( = \frac{9}{21} \) Probability of drawing a red second \( = \frac{4}{21} \) Total probability: \[ P(B \cap R) = \frac{9}{21} \times \frac{4}{21} = \frac{36}{441} = \frac{4}{49} \] ### Without Replacement 5. A red and a red Probability of drawing a red first \( = \frac{4}{21} \) Probability of drawing a red second, without replacement \( = \frac{3}{20} \) Total probability: \[ P(R \cap R) = \frac{4}{21} \times \frac{3}{20} = \frac{12}{420} = \frac{1}{35} \] 6. A red and a yellow Probability of drawing a red first \( = \frac{4}{21} \) Probability of drawing a yellow second, without replacement \( = \frac{8}{20} \) Total probability: \[ P(R \cap Y) = \frac{4}{21} \times \frac{8}{20} = \frac{32}{420} = \frac{8}{105} \] 7. A yellow and a yellow Probability of drawing a yellow first \( = \frac{8}{21} \) Probability of drawing a yellow second, without replacement \( = \frac{7}{20} \) Total probability: \[ P(Y \cap Y) = \frac{8}{21} \times \frac{7}{20} = \frac{56}{420} = \frac{2}{15} \] 8. A blue and a red Probability of drawing a blue first \( = \frac{9}{21} \) Probability of drawing a red second, without replacement \( = \frac{4}{20} \) Total probability: \[ P(B \cap R) = \frac{9}{21} \times \frac{4}{20} = \frac{36}{420} = \frac{3}{35} \]