This should be 1.35, sorry if it’s wrong
Answer:
12 possibilities
Step-by-step explanation:
In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.
The same thing occurs in the second urn, as all balls have different labels.
The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).
For the first urn, we have a combination of 4 choose 2:
C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities
For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.
In total we have 6 + 6 = 12 possibilities.
The answer is y=mx+b y=(-2)-(1/2)
