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A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6

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A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 22 Apr 2020, 10:37
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A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3

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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 22 Apr 2020, 12:16
1
1
Total number of possible arrangements of 3 cards = 6*5*4 = 120

Number of ways to select consecutive integers cards in increasing order = 4 [{1,2,3}, {2,3,4}, {3,4,5}, {4,5,6}]

Probability = 4/120 = 1/30



Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 23 Apr 2020, 03:53
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


total ways to select 3 cards from 6 ; 6*5*4 ; 120 ways
and possible set of consective cards ; (1,2,3) ; ( 2,3,4); ( 3,4,5); ( 4,5,6) ; 4
so P = 4/120 ; 1/30
OPTION B
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 21 May 2020, 21:12
1
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 22 May 2020, 01:57
Mck2023 wrote:
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device


Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

ANswer: Option B
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 22 May 2020, 19:31
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device


Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

ANswer: Option B


Hi GMATinsight
Thank you for the reply.

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 22 May 2020, 20:57
Mck2023 wrote:
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device


Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

ANswer: Option B


Hi GMATinsight
Thank you for the reply.

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!


Mck2023

Use Combination when you need to do ONLY SELECTION

Use Permutaton when you need to do SELECTION + ARRANGEMENT
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One-on-One Skype classes l Classroom Coaching l On-demand Quant course l Admissions Consulting

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Click for FREE Demo on VERBAL & QUANT
Our SUCCESS STORIES: From 620 to 760 l Q-42 to Q-49 in 40 days l 590 to 710 + Wharton l
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 30 May 2020, 17:45
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is \(\frac{1}{6}\)
Selecting the 2nd card is \(\frac{1}{5}\)
Selecting the 3rd card is \(\frac{1}{4}\)

P = 4 x \(\frac{1}{6}\frac{1}{5}\frac{1}{4}\)
P = \(\frac{1}{30}\)

Hope it helps.
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 11 Jun 2020, 22:27
Cellchat wrote:
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is \(\frac{1}{6}\)
Selecting the 2nd card is \(\frac{1}{5}\)
Selecting the 3rd card is \(\frac{1}{4}\)

P = 4 x \(\frac{1}{6}\frac{1}{5}\frac{1}{4}\)
P = \(\frac{1}{30}\)

Hope it helps.



Here then you have assumed that the order you have picked up the cards is also consecutive

Which is not the case mentioned in the question. The question says that the cards picked up line up in order afterwards.
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A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 12 Jun 2020, 11:43
Mck2023 wrote:
GMATinsight wrote:
Mck2023 wrote:
Bunuel wrote:
A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?

A. 1/60
B. 1/30
C. 1/20
D. 1/6
E. 1/3


Hi Bunuel
How do we know to use permutation not combination formula to answer this question?
Can you please put some lights!
Thank you!

Posted from my mobile device


Mck2023

Understand that permulation formula includes selection as well as arrangements

Total Outcomes = Permutation of 3 out of 6 digits = 6P3 = 6! / (6-3)! = 120

Favorable outcomes = {123} or {234} or {345} or {456} = 4 outcomes

Probability = 4/120 = 1/30

ANswer: Option B


Hi GMATinsight
Thank you for the reply.

I am still not clear what in the question make us use permutation formula but nit the combination one?

Thank you!


Permutation is used, when ORDER MATTERS.
In this question, we need to use PERMUTATION, because we need probability of cards to be in ascending order, after we draw first card.
e.g if first card drawn is 3, 2nd card should be 4 - not any other, So order matters. I hope you understood.
Total cases formed are thus 4 / 6p3 : 4/120 = 1/30

Combination is used when order doesn't matter and you have to just find total combination or ways of arrangement of anything.


Kudos if it helped !!!
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 14 Jun 2020, 04:19
1st number from the has to be amongst {1,2,3,4}
Therefore probability of 1st number is 4/6.

2nd number has to be only one of the balance five numbers I.e. next consecutive number. Therefore probability of second number is 1/5

Likewise probability of third number will be 1/4.

So,final answer I.e. probability of three consecutive numbers among given six numbers in increasing order will be 4/6 * 1/5 * 1/4 = 1/30

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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6  [#permalink]

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New post 14 Jun 2020, 13:52
PrajwalGupta wrote:
Cellchat wrote:
To select consecutive integers in increasing order, the possible sets are as follows:

{1, 2, 3} OR
{2, 3, 4} OR
{3, 4, 5} OR
{4, 5, 6}

Next, we note that the probability of selecting any of the 4 sets above is the same. So we can simply multiply the probability of 1 set by 4.

Selecting the 1st card is \(\frac{1}{6}\)
Selecting the 2nd card is \(\frac{1}{5}\)
Selecting the 3rd card is \(\frac{1}{4}\)

P = 4 x \(\frac{1}{6}\frac{1}{5}\frac{1}{4}\)
P = \(\frac{1}{30}\)

Hope it helps.



Here then you have assumed that the order you have picked up the cards is also consecutive

Which is not the case mentioned in the question. The question says that the cards picked up line up in order afterwards.

The question states "A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6. If three cards are randomly selected from the deck without replacement, what is the probability that the numbers on the cards are consecutive integers in increasing order?"

You must pick the cards up in increasing order while at the same time pick consecutively.

There's 4 different ways you can do this, each of which has a probability of \(\frac{1}{120}\).
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Re: A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6   [#permalink] 14 Jun 2020, 13:52

A deck of cards contains 6 cards numbered from 1, 2, 3, 4, 5, and to 6

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