Question

A card is chosen from a standard deck of cards. What is the probability that the card is a king, given that the card is black ? A. 1/26 B. 1/13 C. 1/2 D.2/13 And I have another question. A drawer contains 12 brown socks, 8 white socks, and 2 blue socks. One sock is selected at random. What is the probability that the sock is blue, given that it is NOT brown? A. 1/4 B. 1/5 C. 1/10 D. 1/11. Thank you sorry i'm a noob at this.

Answer 1

EVERY probability question uses this formula.
You should memorize it.

Probability =
                         (number of ways it can happen the way you want it)
             divided by
                                 (total number of ways it can happen).

When a card is chosen from a standard deck of cards, it can turn out
in 52 different total ways.
If you're looking for a black king, there are 2 of those.
So the probability is

                                          2 / 52  =  1 / 26  =  about 3.85%

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In the sock drawer . . .

-- There are 22 socks altogether.

-- What is the total number of ways you can draw a single sock ?
Normally it would be 22.  But the question says the one you draw is
definitely NOT brown.  So you can just toss the brown socks out of
the question entirely.  You're actually working with the total number
of ways you can draw one sock that is not brown.  There are 10 of those.

-- 2 of them are blue.

-- So the probability is . . .

                         2 / 10 = 1 / 5  =  20% .

Answer 2

There are 26 black cards in a deck of cards.

There is a King Of Clubs and a King of Spades.

So, given that the card is black, the chances of it being a King is: 

2/26 which is 1/13.

This can be proven:

P(Card Is Black (B)) = 26/52

P(Card Is King (K)) = 4/52

P(Card Is King And Black) = 2/52

$P\left( K|B \right) =\frac { P\left( K\cap B \right) }{ P\left( B \right) } \\ \\ =\frac { \frac { 2 }{ 52 } }{ \frac { 26 }{ 52 } } \\ \\ =\frac { 2 }{ 52 } \times \frac { 52 }{ 26 } \\ \\ =\frac { 2 }{ 26 } \\ \\ =\frac { 1 }{ 13 }$

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Using Bayes' theorem and:

P(Brown Sock (B)) = 12/22

P(White Sock (W)) = 8/22

P(Blue Sock (L)) = 2/22

P(Not Brown Sock (N)) = 10/22

Had to use this to make sure I got the right answer. Second question was a tad confusing.

$P\left( L|N \right) =\frac { P\left( L \right) \times P\left( N|L \right) }{ P\left( N \right) } \\ \\ =\frac { \frac { 2 }{ 22 } \times 1 }{ \frac { 10 }{ 22 } } \\ \\ =\frac { 2 }{ 22 } \times \frac { 22 }{ 10 } \\ \\ =\frac { 2 }{ 10 } \\ \\ =\frac { 1 }{ 5 }$