Question from Julie, a parent:
Hello,
I am a parent, teacher, student doing an online class to get certified to teach. I have the anwsers to these problems and would like to check to see if they are correct. I already did the work. Just checking if they are right or wrong.
 Select a card from the deck(card deck of 52 cards). Place the card back in the deck. What is the probability that both cards will be black? Is it 25/104, 24%?
 Select a card from the deck. Do not replace the card. Select another card. What is the probability that a red and a face card are the two cards you selected? Is it 2/221, .9%?
 Remove the red cards from the deck and assume that the remaining cards have been shuffled: select a card from the remaining deck. Place the card back in the deck. Select another card. What is the probability that both cards will be black? Is it 26/26 x 26/26= 1 x 1 = 1 , 100%?
Select a card from the remaining deck. Do not replace the card. Select another card. What is the probability that a red 3 and a face card are the 2 cards you selected? Is it 0/26 x 6/26 = 0, 0%?
 Seperate 12 face cards from the rest of the deck. Assume that the face cards have been shsuffled. Select 3 cards from the pile of face cards. How many different arrangements can be made by selecting 1 of each card (jack, king, queen) from the pile? is it 1/4 x 1/4 x 1/4 x 1/4 = 1/64?
 How many different ways are there of selecting the queen of clubs, then the king of diamonds, and then the jack of hearts from the pile? Is it 1/12 x 1/12 x 1/12 = 1/1728?
I have spent a lot of time on this and am not asking for the answers or for you to do my work. Just want to know if they are correct or not so I can go back and fix it if necessary. Hopefully they are correct or on the right track.


Hi Julie,
 Select a card from the deck(card deck of 52 cards).
I am assuming that there is a missing line here "Shuffle and select another card."
Place the card back in the deck. What is the probability that both cards will be black? Is it 25/104, 24%?
No. I think you have made more than one mistake but I'm not sure what.
 Did you get 25/104 as (13 + 12)/(52+52)? If so, the operation of computing
(numerator + numerator)/(denominator + denominator)
is almost NEVER the right answer to ANY question. It is sometimes used as a way to produce a fraction between two others.
 The probability that a card drawn from a full deck is black is 26/52, or 1/2. After you replace & shuffle it is 1/2 again, and independent of the first one. It does not go down (26>25 or 13>12) If it did the denominator would change too.
 To find P(A and B) where A and B are independent events, multiply their individual probabilities.
 Select a card from the deck. Do not replace the card. Select another card. What is the probability that a red and a face card are the two cards you selected? Is it 2/221, .9%?
No. Whether you replace the card or not is irrelevant here, because the events (red) and (face) are independent. Using this insight you can answer the question easily:
P = P(1st red) * P(2nd face)
= P(red) * p(face) for draws from two separate packs
= P(red and face) for a single draw.
Note that this would NOT be true if the two events were not (red, face) but (say) (red, red), (red, black), (face, ace) or (face, king) because these are not independent.
 Remove the red cards from the deck and assume that the remaining cards have been shuffled: select a card from the remaining deck. Place the card back in the deck. Select another card. What is the probability that both cards will be black? Is it 26/26 x 26/26= 1 x 1 = 1 , 100%?
The cards in the remaining deck are all black. A 100% probability represents certainty. Is it certain that both will be black?
Select a card from the remaining deck. Do not replace the card. Select another card. What is the probability that a red 3 and a face card are the 2 cards you selected? Is it 0/26 x 6/26 = 0, 0%?
A 0% probability represents impossibility. Is drawing a red 3 from the remaining deck impossible?
 Seperate 12 face cards from the rest of the deck. Assume that the face cards have been shsuffled. Select 3 cards from the pile of face cards. How many different arrangements can be made by selecting 1 of each card (jack, king, queen) from the pile? is it 1/4 x 1/4 x 1/4 x 1/4 = 1/64?
You are asked for a number of arrangements, which clearly cannot be 1/64; it must be in {0,1,2,3,..} [As no probability is requested, shuffling is irrelevant!]
How many choices for the jack have you got? How many for the queen? The king?
How many for the (J,Q,K) triple?
 How many different ways are there of selecting the queen of clubs, then the king of diamonds, and then the jack of hearts from the pile? Is it 1/12 x 1/12 x 1/12 = 1/1728?
Again, the question  as you have stated it  is "how many ways" and the answer is obvious.
IF you mean "what is the probability" then you have to use the product rule for a conjunction ("and") of dependent events because the probability of getting the KD on the second draw depends on whether you got the QC on the first.
The rule is
P(A&B&C&...K) = P(A) * P(B given A) * P(C given A&B) * ... *P(K given A&B&...&J)
P(QC1) = 1/52.
P(KD2 given QC1) = 1/51 because there are 51 cards left and we know the KD is one of them. Note P(QC2 given QC1) = 0 because there are 51 cards left and we know the QC is NOT one of them.)
And so on...
Good Hunting!
RD
