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Two cards are drawn successively from a standard deck of 52 [#permalink]
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08 Jan 2010, 17:29
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Two cards are drawn successively from a standard deck of 52 wellshuffled cards. Find the probability that a. both are spades b. the 1st card is a jack and the 2ns is not c. the 1st card is a 7 and the 2nd is a 5 of spades d. the 2nd is a face card
good one to clear concepts completely for card questions.



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Re: Probability very good question set [#permalink]
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08 Jan 2010, 20:23
a) P(both are spades)=\(\frac{13C2}{52C2}\) =\(\frac{13*6}{51*26}\) =\(\frac{1}{27}\)
b) P(1st card jack 2nd card is not) = \(\frac{4C1*48C1}{52C2}\) = \(\frac{4*48}{51*26}\) = \(\frac{96}{663}\)
c) P(1st card 7 2nd card 5 of spades) = \(\frac{4C1*1}{52C2}\) = \(\frac{4*1}{51*26}\) = \(\frac{4}{663}\)
d) P(2nd card face card) = \(\frac{(first card non face + both cards face)}{52C2}\) = \(\frac{40*12+12*11}{51*26}\) = \(\frac{306}{663}\) = \(\frac{102}{221}\)



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Re: Probability very good question set [#permalink]
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09 Jan 2010, 00:26
mrblack wrote: a) P(both are spades)=\(\frac{13C2}{52C2}\) =\(\frac{13*6}{51*26}\) =\(\frac{1}{27}\)
b) P(1st card jack 2nd card is not) = \(\frac{4C1*48C1}{52C2}\) = \(\frac{4*48}{51*26}\) = \(\frac{96}{663}\)
c) P(1st card 7 2nd card 5 of spades) = \(\frac{4C1*1}{52C2}\) = \(\frac{4*1}{51*26}\) = \(\frac{4}{663}\)
d) P(2nd card face card) = \(\frac{(first card non face + both cards face)}{52C2}\) = \(\frac{40*12+12*11}{51*26}\) = \(\frac{306}{663}\) = \(\frac{102}{221}\) Two cards are drawn successively from a standard deck of 52 wellshuffled cards. Find the probability that (A) Both are spades: \(\frac{13}{52}*\frac{12}{51}\) (B) The 1st card is a jack and the 2nd is not : \(\frac{4}{52}*\frac{48}{51}\) (C) The 1st card is a 7 and the 2nd is a 5 of spades: \(\frac{4}{52}*\frac{1}{51}\) (D) The 2nd is a face card: \(\frac{12}{52}*\frac{11}{51}+\frac{40}{52}*\frac{12}{51}\) I think the quoted post is not correct as in denominator is 52C2 and there should be 52P2, as order matters in this case: JK is different from KJ.
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Re: Probability very good question set [#permalink]
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09 Jan 2010, 08:15
Why would order matter? For the denominator you just want the # of combinations of getting 2 cards.



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Re: Probability very good question set [#permalink]
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09 Jan 2010, 10:44
mrblack wrote: Why would order matter? For the denominator you just want the # of combinations of getting 2 cards. You are counting the # of selections of TWO cards in denominator in ALL four cases, but in nominator you are counting # of selections of first card and the second card separately, in all cases but the first. So you got correct answers in the first example and wrong in all other examples. Consider this: There are four marbles in the jar: white, black, red and blue. Two marbles are drawn from it. Find the probability that first marble is white and second is red: Now, according to your solution it would be 1C1*1C1/4C2=1/6, which is not correct. Correct answer is: as order matters, we need sequence WR only: 1/4*1/3=1/12. If it were: find the probability of one marble being white and another red, then both WR and RW would be winning scenarios: P=1/4*1/3+1/4*1/3=1/6.
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Re: Probability very good question set [#permalink]
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09 Jan 2010, 11:31
Bunuel wrote: mrblack wrote: Why would order matter? For the denominator you just want the # of combinations of getting 2 cards. You are counting the # of selections of TWO cards in denominator in ALL four cases, but in nominator you are counting # of selections of first card and the second card separately, in all cases but the first. So you got correct answers in the first example and wrong in all other examples. Consider this: There are four marbles in the jar: white, black, red and blue. Two marbles are drawn from it. Find the probability that first marble is white and second is red: Now, according to your solution it would be 1C1*1C1/4C2=1/6, which is not correct. Correct answer is: as order matters, we need sequence WR only: 1/4*1/3=1/12. If it were: find the probability of one marble being white and another red, then both WR and RW would be winning scenarios: P=1/4*1/3+1/4*1/3=1/6. i completely agree with the above example but plz consider below for the question: the question asks for combinations like 1J,5K or 8Q or for that matter 1st card be a non face card and the 2nd be a face card hence solution would be: P(Non face card in the 1st draw) * P(face card in the 2nd draw) ={C(40,1)/C(52,1)} * {C(12,1)/C(51,1)} is it the above case or do we also have to consider the case in which the 1st card is also a face card and 2nd card is also a face card.?
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Re: Probability very good question set [#permalink]
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09 Jan 2010, 12:04
mojorising800 wrote: i completely agree with the above example but plz consider below for the question:
the question asks for combinations like 1J,5K or 8Q or for that matter 1st card be a non face card and the 2nd be a face card hence solution would be: P(Non face card in the 1st draw) * P(face card in the 2nd draw) ={C(40,1)/C(52,1)} * {C(12,1)/C(51,1)}
is it the above case or do we also have to consider the case in which the 1st card is also a face card and 2nd card is also a face card.? I'm not sure I understood your question... Anyway: Question is: find the probability second card is face card. In our original question (#4) we are asked to determine the probability that the second drawn card is face card, it means that second card must be face card, but the first card may or may not be face card. This can occur in two ways (winning scenarios): First drawn is face card*Second drawn is face card+First drawn is not face card*Second drawn is face card=F/F+NF/F=12/52*11/51+40/52*12/51. Both F/F and NF/F are winning scenarios for us, as in both cases there is face card as second card. If it were: find the probability first card is nonface card and second is face card, then only way it can occur is NF/F: NF/F=40/52*12/51 (exactly as you wrote). No need to add to this probability F/F, as it's not a winning scenario for us. Hope it's clear. Tell me if I answered not the question you were interested in and I'll try to elaborate more.
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Re: Probability very good question set [#permalink]
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09 Jan 2010, 12:19
Bunuel wrote: mojorising800 wrote: i completely agree with the above example but plz consider below for the question:
the question asks for combinations like 1J,5K or 8Q or for that matter 1st card be a non face card and the 2nd be a face card hence solution would be: P(Non face card in the 1st draw) * P(face card in the 2nd draw) ={C(40,1)/C(52,1)} * {C(12,1)/C(51,1)}
is it the above case or do we also have to consider the case in which the 1st card is also a face card and 2nd card is also a face card.? I'm not sure I understood your question... Anyway: Question is: find the probability second card is face card. In our original question (#4) we are asked to determine the probability that the second drawn card is face card, it means that second card must be face card, but the first card may or may not be face card. This can occur in two ways (winning scenarios): First drawn is face card*Second drawn is face card+First drawn is not face card*Second drawn is face card=F/F+NF/F=12/52*11/51+40/52*12/51. Both F/F and NF/F are winning scenarios for us, as in both cases there is face card as second card. If it were: find the probability first card is nonface card and second is face card, then only way it can occur is NF/F: NF/F=40/52*12/51 (exactly as you wrote). No need to add to this probability F/F, as it's not a winning scenario for us. Hope it's clear. Tell me if I answered not the question you were interested in and I'll try to elaborate more. yep!!that was my question exactly!! whether winning scenarios were NF/F only or NF/F+F/F.. doubt cleard!!
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Re: Probability very good question set [#permalink]
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09 Jan 2010, 18:58
Bunuel wrote: You are counting the # of selections of TWO cards in denominator in ALL four cases, but in nominator you are counting # of selections of first card and the second card separately, in all cases but the first. So you got correct answers in the first example and wrong in all other examples.
Consider this: There are four marbles in the jar: white, black, red and blue. Two marbles are drawn from it. Find the probability that first marble is white and second is red:
Now, according to your solution it would be 1C1*1C1/4C2=1/6, which is not correct.
Correct answer is: as order matters, we need sequence WR only: 1/4*1/3=1/12.
If it were: find the probability of one marble being white and another red, then both WR and RW would be winning scenarios: P=1/4*1/3+1/4*1/3=1/6. Thanks for the explanation Bunuel.




Re: Probability very good question set
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