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? period______

Explanation:

Step1: Calculate total number of M&Ms initially

First, we sum up the number of each color of M&Ms. The number of red is 5, orange is 4, yellow is 5, purple is 3, and green is 3. So the total number \( n = 5 + 4 + 5 + 3 + 3 \).
\( n = 20 \)

Step2: Probability of first red M&M

The probability of drawing a red M&M first is the number of red M&Ms divided by the total number of M&Ms. Let \( P(R_1) \) be the probability of first red. So \( P(R_1)=\frac{5}{20}=\frac{1}{4} \)

Step3: Probability of second red M&M (after eating first)

After eating the first red M&M, the number of red M&Ms left is \( 5 - 1 = 4 \), and the total number of M&Ms left is \( 20 - 1 = 19 \). Let \( P(R_2|R_1) \) be the probability of second red given first red. So \( P(R_2|R_1)=\frac{4}{19} \)

Step4: Probability of two reds in a row

Since the two events are dependent (because we eat the first M&M, so the total number and number of reds change), we use the multiplication rule for dependent events: \( P(R_1 \cap R_2)=P(R_1)\times P(R_2|R_1) \)
Substituting the values, we get \( P(R_1 \cap R_2)=\frac{1}{4}\times\frac{4}{19}=\frac{1}{19} \)

Answer:

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