Answer:
A) The best way to picture this problem is with a probability tree, with two steps.
The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.
The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.
If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.
If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.
B) So a person can have a red marble and a blue marble in two ways:
1) Picking the red first and the blue last
2) Picking the blue first and the red last
C) P(R&B) = 3/5 = 60%
Step-by-step explanation:
C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5
For me personally, the easiest way to do this is by isolating the x² term, and finding the square root of both sides. The hardest way (well actually, the longest way) would be to use the quadratic formula. It just complicates things unnecessarily.
