Nitai M. answered • 12/08/19

Experienced High School and College Tutor Specializing in STEM

Let's begin by modeling the procedure described. This is the key to solving any problem in probability. If we can understand in words what is being done, then we can convert our description into a mathematical representation and solve.

PART 1:

One card is removed at random from the deck. This may be a black king or it may be any other card. (Since our goal in the end is to draw a black king, it will be important to know what card is drawn so we know how many black kings remain during the second step. We obviously can't know what card is drawn, we must consider both possibilities).

PART 2: Fifty-one cards remain and a card is picked. It may be a black king, or it may be any other card.

PUTTING IT ALL TOGETHER

Picking a black king in the end can happen in one of two ways:

(First draw = BLACK KING) __AND__ (Second draw = BLACK KING)

__OR__

(First draw = **NOT** BLACK KING) __AND__ (Second draw = BLACK KING)

The problem contains 4 situations (each in its own parentheses above) who's probabilities we must model and connect appropriately. Let's model them first and then we will connect them.

__Probability of (First draw = BLACK KING):__

In the **numerator** we want to write the number of ways our special case can occur. Specifically, how many ways are there in which a black king can be drawn from a deck of 52 cards. **Two**! (There are two black kings in each deck.)

In the **denominator** we will write the number of ways there are of drawing the first card. **Fifty-two**! (There are 52 cards we could pick.)

So, P(First draw = BLACK KING) = __2/52__

__Probability of (Second draw = BLACK KING)__, given (First draw = BLACK KING):

We are only left with **one** black king in the deck, and the deck now has only 51 cards.

Therefore, P(Second draw = BLACK KING) = __1/51__

__Probability of (First draw = ____NOT____ BLACK KING):__

In this case, we have **50** cards we can choose that are not a black king, and **52** possible ways of choosing one card. So,

P(First draw = **NOT** BLACK KING) = __50/52__

__Probability of (Second draw = BLACK KING)__, given (First draw = NOT BLACK KING):

In this case, both black kings are still in the deck (**2** ways of picking a black king) and there are only **51** total cards remaining in the deck (51 possible ways of choosing one card).

P(Second draw = BLACK KING) = __2/51__

CONNECTING ALL THE PIECES:

Whenever we say __AND__ in our model, we should think __multiplication__.

Whenever we say __OR__ in our model, we should think __addition__.

So, we have for the probability of picking a black card from a deck after throwing out one card at random:

P = (2/52)*(1/51) + (50/52)*(2/51)

Now just evaluate the model and you're done!

P = 0.0385

Hope this helps! I know it's a long answer, but hopefully talking out the whole process helps you with future problems like this one.