basic probability question counter intuitive?

[BruceZ]

[ QUOTE ]
Note that

P = A + A'B + A'B'C

= A + A'(B + B'C)

The latter is the method given by BruceZ where one multiplication less gets performed.

[/ QUOTE ]

Mine isn't the factored version. It is the original version with A'B' implemented by the equivalent 1 - (A + A'B) where we already computed (A + A'B).

The factored form says "Hit on the 1st card OR (miss on the first card AND hit on one of the other 2 cards)". My version says "hit on one of the first 2 cards OR (miss on the first 2 cards AND hit on the 3rd card)".

[R Gibert]

Okay, you're exactly right. What I've said about your formula was wrong.

What you have is literally:

P = (A + A'B) + (1 - (A + A'B))C

and how you describe it is exactly right:

= A + A'B + (1 - A - A'B)C
= A + A'B + (A' - A'B)C
= A + A'B + A'(1 - B)C
= A + A'B + A'B'C

where

A'B' = 1 - (A + A'B)

just as you said.

[arden street]

thank you for the detailed replies