Question

a bag of m&ms has 3 red, 5 green, 4 blue, and 2 yellow m&ms. suppose you randomly select two m&ms from the bag one at a time without replacing the first m&m. let a = first m&m is red and b = second m&m is green. find the following probabilities. (write your answers as fractions.) a) p(a) = b) p(b | a) = c) p(a and b) = question help: message instructor

Explanation:

Step1: Calculate total number of M&M's

Total M&M's = 3 (red) + 5 (green) + 4 (blue) + 2 (yellow)=14.

Step2: Calculate P(A)

The probability that the first M&M is red is the number of red M&M's divided by the total number of M&M's. So \(P(A)=\frac{3}{14}\).

Step3: Calculate P(B|A)

If the first M&M is red (event A occurs), then there are 13 M&M's left. The probability that the second M&M is green given that the first is red is the number of green M&M's divided by the remaining number of M&M's. So \(P(B|A)=\frac{5}{13}\).

Step4: Calculate P(A and B)

By the formula \(P(A\cap B)=P(A)\times P(B|A)\), we substitute the values from previous steps: \(P(A\cap B)=\frac{3}{14}\times\frac{5}{13}=\frac{15}{182}\).

Answer:

a) \(\frac{3}{14}\)
b) \(\frac{5}{13}\)
c) \(\frac{15}{182}\)