Playing cards word problem

[MathLearner2016]

5 cards are drawn from a normal deck of cards.
a) What are the odds that drawn was 1 five card and 2 three cards?
b) What are the odds that 3 of the cards were of the same suit?

[thatsneakyguy]

What do you think the answers are?

 

How far have you gotten with these problems?

[Country Boy]

"5 cards are drawn from a normal deck of cards.
a) What are the odds that drawn was 1 five card and 2 three cards?"

​I presume you mean "exactly 1 five card and 2 three cards" and not "at least that many".

 

​ Can we at least assume you know that there are 52 cards in a "normal deck of cards", 13 in each suit, and 4 of each rank?

 

​ There are 4 "5" cards so the probability of drawing a "5" on the first draw is 4/52= 1/13. There are then 51 cards left in the deck, 4 of them "3"s. The probability of drawing a "3" on the second draw is 4/51. There are then 50 cards left in the deck, 3 of them "3"s. The probability of a "3" on the next draw is 3/50. There are then 49 cards left in the deck, three of them "5"s and two of them "3"s so there are 52- 3- 2= 47 of them not "5" or "3". The probability of drawing any card other than a "3" or a "5" is 47/49. There are then 48 cards left in the deck, 46 of them neither "5" or "3". The probability the last card drawn is 46/48. Therefore the probability drawing "5", "3", "3", "A", "A", where "A" is "any card other than a 3 or a 5", is (4/52)(4/51)(3/50)(47/49)(6/48).

 

Do the same argument for "5, A, 3, A, 3" or any other order of the same 5 cards to see that the probability of drawing a "5", two "3"s, and two other cards is the same (you have different fractions but just with the numerators rearranged so the product is the same). There are [tex]\frac{5!}{1!2!2!}= \frac{120}{4}= 30[/tex] different arrangements of a "5", two "3"s, and 2 other cards so multiply that previous probability by 30 to get the probability of a "5", two "3"s, and two of any other cards.

 

"b) What are the odds that 3 of the cards were of the same suit?"

​ I presume you mean exactly "3 cards of the same suit" and not "at least 3 cards of the same suit".

 

​The first card can be of any suit. That leaves 51 cards, 12 of the same suit as the previous card so the probability the second card is of the same suit as the first is 12/51. That leaves 50 cards, 11 of the same suit as the two previous cards so the probability the third card is also of that suit is 11/50. That leaves 49 cards, 39 of which are not of the same suit as the previous three. The probability that the fourth card is of a different suit from the first three is 39/49. That leaves 48 cards, 38 of which are not of the same suit as the first three (but might be of the same suit as the fourth). The probability that the fifth card is not of the same suits as the first three (but might be of the same suit as the fourth) is 38/48. The probability of getting the first three cards of the same suit and the last two of other suits is (12/51)(11/50)(39/49)(38/48). There are [tex]\frac{5!}{3!1!1!}= \frac{120}{6}= 20[/tex] different rearrangements that has three cards of the same suit so we need to multiply by 20 to get the probability of "three cards of the same suit" in any order.