#4
White block(s) = 2
Red block(s) = 1
Purple block(s) = 3
Total = 2 +1 + 3 = 6 blocks
a) P(white) = [tex] \frac{2}{6} = \frac{1}{3} [/tex]
P(red) = [tex] \frac{1}{6} [/tex]
P(purple) = [tex] \frac{3}{6} = \frac{1}{2} [/tex]
b)Not white block:
1 - [tex] \frac{1}{3} [/tex] OR [tex] \frac{1}{2} + \frac{1}{6} [/tex]
[tex] \frac{2}{3} [/tex] [tex] \frac{2}{3} [/tex]
Because, when they say no white blocks, we simply do not count them and add the rest to find that probability without white blocks.
c) The probability stays the same: lets say now we have
4 white blocks, 2 red, and 6 purple, total will be 12
P(white)=[tex] \frac{4}{12} [/tex] which is still [tex] \frac{1}{3} [/tex]
d) We get two more blocks in the numerator: lets say we have 4 white blocks, 3 red, 5 purple (after adding 2 of each color), total will be 12
P(purple)=[tex] \frac{5}{12} [/tex]
(im not quite sure if my explanation here helps you though)
e) 1 more of white and purple, 5 more of red
white = 3, purple = 4, red = 6, total = 12
(you can either add 2 to white or purple but make sure you add 5 of red)
P(red)=[tex] \frac{6}{12} [/tex] = [tex] \frac{1}{2} [/tex]
