Question

32 cards, contains a) One scendon Gate Solution of the exercise 1.3 The sample space is given by the s tancos choices in the same time of 13 cards from 52, then -------Pis the form probability on n ews the choices are of same probability The event A: Oneace and one cely s by choosing the 4 aces, then 12 other cards from the others, the A- obtain then the probability: P(A) - - For the event B:"The band contains one ace at least", we count the probability of B: "The hand doesn't contain any sce", which is realized by choosing 13 from the cards different of aces, then: B ........., which gives the pea ky P(B) - 3 and then the wanted probability: P(B) = 1 - P(B) = .....

Answer 1

= selecting 13 cards from 52 cards. (2)=112) = Borse number of way selecting 13 cards from 52 is c(52,13) ie 11=526=52! - 52! 131(52–13)! 13/39! A: One ace and onlq ace only” is realized by choosing one ace from aces, then 12 other cards from other cards. (Al = 46,.52-46. = 46.486,2 = 4.48! I The Ladeof ways 121 36! conve rtirse en una selection of 12 from one ace jmémwning (52-5) from 4 ace cards. PCA) = \AL = 4.48! Tel 12736 52! = 4481031391 2536! 521 Elix 481x13x121x39x38 X 37 361 121*36) * 52X51X50x49 *481 =4x13x39x38 X37 52x51 X50X48 . = luxxx 29 x 33 x 37 2851368 6364800 = 0.4479 52x51x50x58 17 12 I PA=0.45

B: The hand does not contain any ace", which es realised by choosing 13 cards from the Cards different of aces, ..(B1: 400. 52-463 = 1.84812 = 486 13135) & P(B) =( 481 13135) 48Į x B 131 x 39! 13! *351 x 52 (521 (13) 391 = 39x38x37 x36x35! x48 351 x 52x 51 x 50 x 49% 481 = 39 x 38 x 37 x36 = 1974024 52 x 51 X 50 X496497400 = 0.303817 1 .:p(B)=0 304 )