How to correctly apply probability multiplication and addition laws?

Question

I'm trying to apply the probability addition rule to the problem below.

There are 12 different socks in a drawer. The table below shows the different varieties:

Thickness	chunky (C) or thin (T)
Style	stripy (S) or dotty (D) or plain (P)
Colour	red (R) or blue (B)

Thickness	Style	Colour
C	S	R
C	S	B
C	D	R
C	D	B
C	P	R
C	P	B
T	S	R
T	S	B
T	D	R
T	D	B
T	P	R
T	P	B

Based on the table, some simple observations:

• Probability that a chunky sock is taken out: 6:12
• Probability that a stripy red sock is taken out: 2:12

This is where i'm getting confused based on applying the laws:

• Probability that a dotty and red sock is taken out:

  • probability of dotty sock = 4:12
  • probability of red sock = 6:12
  • applying multiplication law, probability of dotty and red sock = 4/12 * 6/12 = 1:6
  • 1:6 seems to correctly reflect observed data in table, so I assume multiplication law is correctly applied in this case?

• Probability that a sock that is neither plain nor blue is taken out:

  • probability of plain sock = 4:12
  • probability of blue sock = 6:12
  • applying addition law, probability of plain or blue sock = 4/12 + 6/12 = 10:12
  • therefore probability of neither plain or blue sock is everything else i.e 2:12 = 1:6
  • observed data in table suggest this should be 4:12 = 1:3
  • What might be wrong in my understanding of the problem and/or application of the addition law?

Answer 1

The probability that a dotty and red sock is taken is 1:6 is correct.
The mistake in the second method:
Let A be one event and B be the second event.
Neither A nor B means (not A) and (not B)
The probability that neither A nor B is selected is $P($not $A) \cdot P($not $B)$
In your case,
Probability that a sock that is neither plain nor blue is taken out = $P($not blue$) \cdot P($not plain$)$
P(not blue)= $1 - \frac{6}{12} = \frac{1}{2}$
P(not plain) = $1 - \frac{4}{12} = \frac{2}{3}$

Probability that a sock that is neither plain nor blue is taken out = $\frac{1}{3}$

Hope this helps
EDIT:
P(A or B) = P(A) + P(B) - P(A and B)
P(A and B) = P(A).P(B) only when A and B are independent. Independent means that an effect on A does not affect B.

Basically
P(neither A nor B) = 1- P(A or B) = 1 - P(A) - P(B) +P(A and B)
Now in this question, A and B are independent so P(A and B) = P(A)P(B)
So,
P(neither A nor B) = 1- P(A or B) = 1 - P(A) - P(B) +P(A)P(B)
$---------------------------------------$ Also,
P(neither A nor B) = not(P(A)) and not (P(B))
So,
P(neither A nor B) = (1 - P(A))(1 - P(B)) = 1 - P(A) - P(B) +P(A)P(B)
You get the same result in both cases.
If you have any more doubts you can ask in the comment