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# A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are

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Intern
Joined: 17 Dec 2017
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Location: Uzbekistan
Schools: Wharton '20
GMAT 1: 650 Q39 V40
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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, 20:54
1
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:35) correct 70% (02:16) wrong based on 42 sessions

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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

Originally posted by NewGuy12 on 05 Feb 2018, 09:53.
Last edited by Bunuel on 08 Feb 2018, 20:54, 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, 01:10
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 re-check 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, 20:44
2
1
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
_________________

Karishma
Veritas Prep GMAT Instructor

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Joined: 01 Jun 2015
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Location: India
GMAT 1: 620 Q48 V26
Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are  [#permalink]

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12 Feb 2018, 10:19
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

VeritasPrepKarishma

Maam thanks for your great explanation.But will the real gmat want us to play with that big numbers??
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8880
Location: Pune, India
Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are  [#permalink]

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12 Feb 2018, 20:56
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

VeritasPrepKarishma

Maam 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.
_________________

Karishma
Veritas Prep GMAT Instructor

Re: A bag contains 50 tickets numbered 1, 2, 3, 4.....50 of which five are   [#permalink] 12 Feb 2018, 20:56
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