Find the probability of getting a full house(3 cards of one denomination and 2 of another) when 5 cards are de

Question

Find the probability of getting a full house(3 cards of one denomination and 2 of another) when 5 cards are dealt from an ordinary deck.

Answer 1

There are 13 ways to choose the first denomination and 12 to choose the second. [Or P(13, 2) = 13×12 ways to choose them both - order is important here.] There are C(4, 3) = 4 ways to choose three cards from the first denomination and C(4, 2) = 6 ways to choose two cards from the second.

So there are 13(12)(4)(6) = 3744 possible ways to get a full house, and so the probability of getting a full house is 3744 / C(52, 5) = 1.44×10^(-3).

Answer 2

nicely there are fifty two enjoying cards interior the deck. The danger of choosing any card is obviously a million/a million. yet then your 2d card would desire to be the same because of the fact the 1st card, so which you have a three/fifty one or a million/17 danger of choosing the same card. as quickly as extra, you may %. that card back, yet this time you basically have a 2/50 or a million/25 shot of having it. So appropriate now, we've a million/a million*a million/17*a million/25 for in simple terms getting 3 of a style. in spite of the undeniable fact that, you nonetheless choose 2 extra enjoying cards of the same fee, the possibility of that's a million/a million*a million/sixteen = a million/sixteen. finally, multiply all of it mutually to get a million/6,800