A bag contains 8 red, 12 white and 4 blue marbles. If 2 marbles are selected in succession (without replacement)

simplest-form
common-fraction

#1

A bag contains 8 red, 12 white and 4 blue marbles. If 2 marbles are selected in succession (without replacement), find the probability as a common fraction in simplest form that: ___a.)both are blue b.)neither is red c.)either is red or blue d.)first is red and second is blue___e.)second is orange

Answer:

Sorry for the rather paltry presentation compared to the other question I answered for you, but I am way too exhausted to type up 2 huge explanations. However, if you are still curious as to the approach feel free to send me a message over Slader or start a session with me at some point in the future and I will do my best to clarify what’s going on here!

A) Probability of drawing the first blue marble = 4/24 = 1/6
Probability of drawing the second blue marble = 3/23
Probability that both marbles are blue = P(1st is blue) * P(2nd is blue) = 1/6 * 3/23 = 3/138 = 1/46

B)Probability first marble is NOT red = 16/24 = 2/3
Probability second marble is NOT red = 15/24 =5/8
Probability that both drawn marbles are NOT red = same approach as A- multiply the two above probabilities:
2/3 * 5/8 = 10/24 = 5/12

C)Probability first marble is blue OR red = 12/24 = 1/2
Probability second marble is blue or red = 11/23
Probability BOTH marbles are blue or red = same as A and B, multiply the probabilities:
1/2 * 11/23 = 11/46

D) Probability of first marble being red = 8/24 = 1/3
Probability second is blue = 4/23
Probability of first red and second blue = 1/3 * 4/23 = 4/69

E) This probability is ZERO since there are no orange marbles, so it doesn’t matter what we draw first as there’s no orange marble to draw!

NOTE: It’s late, if someone can check my multiplication (and even approach to this as I am very tired), I would greatly appreciate any confirmation or corrections!