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A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are [#permalink]
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Updated on: 08 Feb 2018, 21:54
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A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are drawn at random and are arranged in ascending order of magnitude. Find the probability that third drawn ticket is equal to 30. a) 551/15134 b) 1/2 c) 551/15379 d) 1/9 e) 1/50
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Originally posted by NewGuy12 on 05 Feb 2018, 10:53.
Last edited by Bunuel on 08 Feb 2018, 21:54, edited 1 time in total.
Renamed the topic, edited the question and the OA.



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Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are [#permalink]
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08 Feb 2018, 02:10
NewGuy12 wrote: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are drawn at random and are arranged in ascending order of magnitude. Find the probability that third drawn ticket is equal to 30.
a) 55/15134
b) 1/2
c) 551/15379
d) 1/9
e) 1/50 Hi NewGuy12, Please recheck the Options Again. Option A should be 551 instead of 55. First and second number should be from tickets numbered 1 to 29 = \(29C2\) ways and remaining two in \(20C2\) ways. Therfore,favourable number of events = \(29C2*20C2\) Hence,required probability = \(\frac{29C2*20C2}{50C5}=\frac{551}{15134}\)
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Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are [#permalink]
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08 Feb 2018, 21:44
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NewGuy12 wrote: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are drawn at random and are arranged in ascending order of magnitude. Find the probability that third drawn ticket is equal to 30.
a) 55/15134
b) 1/2
c) 551/15379
d) 1/9
e) 1/50 Since we are picking 5 distinct numbers, there is only one way of arranging them in increasing order so we don't have to worry about the arrangement. We just need to pick the right numbers. To ensure that 30 comes right in the middle, 2 numbers should be smaller than it and two greater. The two smaller numbers can be picked in 29C2 ways out of 1  29. The two greater numbers can be picked in 20C2 ways out of 31  50. The 5 numbers can be chosen out of 50 in 50C5 ways. Hence probability = (29C2 * 20C2) / 50C5 = (29*28*20*19)*5! / (50*49*48*47*46)*2*2 = 551/15134
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Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are [#permalink]
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12 Feb 2018, 11:19
VeritasPrepKarishma wrote: NewGuy12 wrote: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are drawn at random and are arranged in ascending order of magnitude. Find the probability that third drawn ticket is equal to 30.
a) 55/15134
b) 1/2
c) 551/15379
d) 1/9
e) 1/50 Since we are picking 5 distinct numbers, there is only one way of arranging them in increasing order so we don't have to worry about the arrangement. We just need to pick the right numbers. To ensure that 30 comes right in the middle, 2 numbers should be smaller than it and two greater. The two smaller numbers can be picked in 29C2 ways out of 1  29. The two greater numbers can be picked in 20C2 ways out of 31  50. The 5 numbers can be chosen out of 50 in 50C5 ways. Hence probability = (29C2 * 20C2) / 50C5 = (29*28*20*19)*5! / (50*49*48*47*46)*2*2 = 551/15134 VeritasPrepKarishmaMaam thanks for your great explanation.But will the real gmat want us to play with that big numbers??



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Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are [#permalink]
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12 Feb 2018, 21:56
techiesam wrote: VeritasPrepKarishma wrote: NewGuy12 wrote: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are drawn at random and are arranged in ascending order of magnitude. Find the probability that third drawn ticket is equal to 30.
a) 55/15134
b) 1/2
c) 551/15379
d) 1/9
e) 1/50 Since we are picking 5 distinct numbers, there is only one way of arranging them in increasing order so we don't have to worry about the arrangement. We just need to pick the right numbers. To ensure that 30 comes right in the middle, 2 numbers should be smaller than it and two greater. The two smaller numbers can be picked in 29C2 ways out of 1  29. The two greater numbers can be picked in 20C2 ways out of 31  50. The 5 numbers can be chosen out of 50 in 50C5 ways. Hence probability = (29C2 * 20C2) / 50C5 = (29*28*20*19)*5! / (50*49*48*47*46)*2*2 = 551/15134 VeritasPrepKarishmaMaam thanks for your great explanation.But will the real gmat want us to play with that big numbers?? No it won't. You won't have tedious calculations in the actual test. The answer might be in this format (29C2 * 20C2) / 50C5 or the numbers might be much smaller.
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