Question

(question 4 only plz)

Answer 1

#4
White block(s) = 2
Red block(s) = 1
Purple block(s) = 3
Total = 2 +1 + 3 = 6 blocks

a) P(white) = $\frac{2}{6} = \frac{1}{3}$
    P(red) = $\frac{1}{6}$
    P(purple) = $\frac{3}{6} = \frac{1}{2}$

b)Not white block:
   1 - $\frac{1}{3}$   OR   $\frac{1}{2} + \frac{1}{6}$
    $\frac{2}{3}$                 $\frac{2}{3}$
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)=$\frac{4}{12}$ which is still $\frac{1}{3}$

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)=$\frac{5}{12}$
(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)=$\frac{6}{12}$ = $\frac{1}{2}$