invernessgangshow.net->Probability-and-statistics-> SOLUTION: two cards are drawn in sequence without replacement from a traditional deck the 52 cards. What is the probability that the an initial card is a confront card (jack, queen, or king) given that var visible_logon_form_ = false;Log in or register.Username: Password: register in one basic step!.Reset your password if girlfriend forgot it."; return false; } "> log On

You are watching: Two cards are drawn without replacement from a standard deck of 52 cards

Click here to view ALL troubles on Probability-and-statisticsQuestion 1118548: 2 cards are drawn in sequence without instead of from a standard deck of 52 cards. What is the probability the the first card is a challenge card (jack, queen, or king) provided that the 2nd card is one eight? (Round your answer to 3 decimal places.) found 4 remedies by rothauserc, greenestamps, ikleyn, Alan3354:Answer by rothauserc(4717) (Show Source): You have the right to put this solution on her website! let occasion A it is in the probability of first drawing a face card = 12/52 = 3/13:let B be the probability of second drawing an 8 = 4/51(because the challenge card was attracted first):Probability (P) (A|B) = p (A intersection B) / P(B):P (A intersection B) method P that both occasions happen, therefore:P (A intersection B) = (3/13) * (4/51) = 12/663 = 4/221:P (A|B) = (4/221) / (4/51) = 51/221 is roughly 0.231: price by greenestamps(9857) (Show Source): You can put this equipment on your website! The inquiry doesn"t yes, really make any sense. The probability the the first card is a face card is not affected by what card is attracted second.The systems by the various other tutor actually reflects this; however it is difficult to see because of the method he shows his calculations, and also by the fact that his final answer in fraction form is not in most basic form.Let A represent illustration a challenge card on the very first draw and also B represent drawing an 8 on the second. ThenP(A) = 12/52 = 3/13P(B) = 4/51Then the conditional probability isP(A|B) = (P(A and also B)/P(B)) = (P(A)*P(B))/P(B) = P(A) = 3/13 prize by ikleyn(41722) (Show Source): You can put this equipment on her website! .The answer to this inquiry is obvious: the probability under the inquiry is = .The answer come this question is obvious: the probability under the question is = .The answer to this inquiry is obvious: the probability under the inquiry is = .

See more: Aluminum Number Of Valence Electrons In Aluminum Number Of Valence Electrons

Price by Alan3354(67667) (Show Source): You have the right to put this solution on your website! This is the 2nd time cards space drawn, yet among them is "given."Makes no sense.Think about it.