Question

1. erin has a bag of m&m’s™. inside the bag are 5 red, 4 orange, 5 yellow, 3 purple and 3 green m&m’s™. what is the probability of drawing a red m&m™ two times in a row if the first m&m™ is eaten before the second m&m™ is drawn?

1. karen has a fun-size bag of skittles™. inside the bag are 2 red, 4 orange, 2 yellow, 2 purple and 2 green skittles™. what is the probability of drawing an orange skittle™, eating it, and then drawing a purple skittle™?

Explanation:

Problem 8:

Step1: Find total number of M&Ms

First, calculate the total number of M&Ms. There are 5 red, 4 orange, 5 yellow, 3 purple, and 3 green. So total \( n = 5 + 4 + 5 + 3 + 3 = 20 \).

Step2: Probability of first red

The probability of drawing a red M&M first is the number of red M&Ms divided by total, so \( P(\text{first red}) = \frac{5}{20} = \frac{1}{4} \).

Step3: Probability of second red (after eating first)

After eating one red M&M, the number of red M&Ms left is \( 5 - 1 = 4 \), and total M&Ms left is \( 20 - 1 = 19 \). So probability of second red is \( P(\text{second red}) = \frac{4}{19} \).

Step4: Probability of both events

Since the two events are dependent, we multiply the probabilities: \( P(\text{two reds}) = \frac{1}{4} \times \frac{4}{19} = \frac{1}{19} \).

Step1: Find total number of Skittles

First, calculate the total number of Skittles. There are 2 red, 4 orange, 2 yellow, 2 purple, and 2 green. So total \( n = 2 + 4 + 2 + 2 + 2 = 12 \).

Step2: Probability of orange first

The probability of drawing an orange Skittle first is the number of orange Skittles divided by total, so \( P(\text{orange}) = \frac{4}{12} = \frac{1}{3} \).

Step3: Probability of purple after eating orange

After eating one orange Skittle, the total number of Skittles left is \( 12 - 1 = 11 \), and the number of purple Skittles is still 2. So probability of purple is \( P(\text{purple}) = \frac{2}{11} \).

Step4: Probability of both events

Multiply the probabilities of the two dependent events: \( P(\text{orange then purple}) = \frac{1}{3} \times \frac{2}{11} = \frac{2}{33} \).

Answer:

\(\frac{1}{19}\)

Problem 9: