Question

Suppose two M&M's are drawn from a bag with 4 green M&M's, 1 brown M&M's, 2 red M&M's, 2 orange M&M's, and 3 blue M&M's. The first M&M is not replaced before the second M&M is drawn. Find each probability below:

a) P(red then green)
b) P(blue then blue)

Answer 1

To find the probability of drawing two M&M's from the bag without replacement, we multiply the probabilities of each event happening. For P(red then green), the probability is 8/45. For P(blue then blue), the probability is 1/12.

Step-by-step explanation:

To find the probability of drawing two M&M's from the bag without replacement, we need to determine the total number of possible outcomes and the number of desired outcomes.

a) P(red then green):

• There are 2 red M&M's and 4 green M&M's in the bag. So, the probability of drawing a red M&M first is 2/9.
• After drawing a red M&M, there are now 1 red M&M and 4 green M&M's left in the bag, making a total of 5 M&M's to choose from.
• So, the probability of drawing a green M&M second is 4/5.
• To find the probability of both events happening, we multiply the probabilities together: (2/9) * (4/5) = 8/45.

b) P(blue then blue):

• There are 3 blue M&M's in the bag. So, the probability of drawing a blue M&M first is 3/9.
• After drawing a blue M&M, there are now 2 blue M&M's left in the bag, making a total of 8 M&M's to choose from.
• So, the probability of drawing a blue M&M second is 2/8.
• To find the probability of both events happening, we multiply the probabilities together: (3/9) * (2/8) = 1/12.