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805+ (Hard)|   Probability|      
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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five Updated on: 12 May 2024, 00:54
Originally posted by NewGuy12 on 05 Feb 2018, 10:53.
Last edited by Bunuel on 12 May 2024, 00:54, edited 2 times in total.
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95% (hard)

Question Stats:

36% (02:42) correct 64% (02:19) wrong based on 69 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
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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 08 Feb 2018, 02:10
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NewGuy12
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
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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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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 08 Feb 2018, 21:44
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NewGuy12
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
Show more

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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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 12 Feb 2018, 11:19
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VeritasPrepKarishma
NewGuy12
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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VeritasPrepKarishma

Maam thanks for your great explanation.But will the real gmat want us to play with that big numbers??
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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 12 Feb 2018, 21:56
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techiesam
VeritasPrepKarishma
NewGuy12
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??
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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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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 12 May 2024, 00:10
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This is a high quality question
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Regor60
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A bag contains 50 tickets numbered 1, 2, 3, 4, ..., 50 of which five 12 May 2024, 06:10
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This question is poorly worded for the intended answer.

All of the numbers must be drawn before they can be arranged.

The third number drawn obviously occurs before all the numbers are drawn and arranged.

The probability that the third number drawn is 30 is:

1/50

The probability that the third number in the arrangement is 30 is the given answer.

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