There are 10 balls in the Urn Total.
Red: 6
Green: 4
Question One: The probability that five red and two green is selected is likely. (as that is over half for both)
Question Two: Impossible. There is only 6 red balls, and 7 are taken from the urn. Thus it would at most be possible for 6 red and 1 green.
Question Three: At least four is likely, as there is more red then green in the Urn.
Hope I helped!
(Mark Brainliest if you can please!)
Answer:
y^2+y-12
Step-by-step explanation:
Once again, FOIL is the way to go!
First, Outside, Inside, Last
y^2+y-12
Than answer you looking for is I think B
45 because 1-4 round down and 6-9 round up to make the 44 become a 50
Answer:
A
Step-by-step explanation:
