What is the probability that it is both a king and a club?

[ilikephysics]

Question: A card is selected from a shuffled deck of cards. What is the probability that it is either a king, or a club? What is the probability that it is both a king and a club?

Answer:
There are 52 cards in a deck. Four kings and 13 clubs. So the probability of chosing a king or a club is 17/52. The probability that it's both a king and a club is 1/52 because it's only one king that's a club. Am I right?


Question:
Five cards are dealt from a shuffled deck. What is the probability that they are all of the same suit? That they are all diamonds? That they are all face cards? That the five cards are a squence in the same suit (e.g. 3,4,5,6,7 of hearts)?

Answer: I need help with the same suit and squence questions. I also need to know if I'm right about the others

Same suit? Not sure about this one
all diamonds? 5/13
face cards? 5/12 (4 kings, 4 queens, 4 jacks)
a squence in same suit? I don't know how to do this one

 

[HallsofIvy]

For the first question, you are almost correct- king OR club: 16/52. One of the kings is a club- you can't count it twice.
king and club: 1/52.

You could also have done the first part by using the formula:
P(A or B)= P(A)+ P(B)- P(A and B)
= 13/52+ 4/52- 1/52= 16/52.

Probability of all 5 cards same suit? One way to do that is this: Since the suit doesn't matter, the first card could be anything but the second card MUST be of that suit: There are 52 cards, 13 cards in a suit. Since you already have one card of the specific suit, there are 51 cards left, 12 of that suit. probability that second card will be same suit as first: 12/51. Now there are 50 cards left, 11 of that suit: probability 3rd card will be of same suit: 11/50. For fourth and fifth cards, same thing: 10/49 and 9/48. Probability all 5 cards of same suit: (12/51)(11/50)(10/49)(9/48)= (12*11*10*9)/(51*50*49*48).

The probability that 5 cards are all diamonds is not nearly as large as 5/13! You could either do the same as above but say: probability first card is a diamond is 13/51 (equals 1/4) and then: 12/51, 11/50, etc. but an easier way is simply to say: there are 4 suits so the probability that suit you got is diamonds is 1/4. Now multiply by the answer to the "any suit": (1/4)(12*11*10*9)/(51*50*49*48).

The probability that the cards are all face cards: there are 12 face cards (Jack, Queen, King in each suit) so the probability that the first card is a face card is 12/52. If that first is a face card, then there are 11 face cards left amoung the remaining 51 cards: probability that second card is also a face card is 11/51. Similarly, for third, fourth fifth cards, the probabilities are 10/50, 9/49, and 8/48. The probability of getting 5 face cards is (12*11*10*9*8)/(52*51*50*49*48).

In order to get a straight, the lowest card cannot be Jack or higher: Jack, Queen, King, Ace of any suit: 16 leaving 52-16= 36 cards. Probability first card will be lower than Jack is 36/52. Now, probability next card is THE next card in the same suit is 1/52. Probability third card is next in that same suit is 1/52, same for fourth and fifth cards. Probability of a straight flush dealt in that order is (36/52)(1/52)⁴. However, order does not matter so multiply by the number of different orders for 5 cards: 5!. The probability of a straight flush is 5!(36/52)(1/52)⁴.

 

[M98Ranger]

without knowing which suit is being considered. If we consider all suits, the probability would be (4/52)*(4/51)*(4/50)*(4/49)*(4/48) = 1/270,725, if we consider just one suit, the probability would be (4/13)*(3/12)*(2/11)*(1/10)*(0/9) = 0.0001286.