GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Nov 2019, 15:12 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Two cards are drawn successively from a standard deck of 52

Author Message
TAGS:

### Hide Tags

Manager  Joined: 25 Dec 2009
Posts: 71
Two cards are drawn successively from a standard deck of 52  [#permalink]

### Show Tags

1 00:00

Difficulty:

(N/A)

Question Stats: 100% (01:14) correct 0% (00:00) wrong based on 10 sessions

### HideShow timer Statistics

Two cards are drawn successively from a standard deck of 52 well-shuffled cards. Find the probability that
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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59095
Re: Probability very good question set  [#permalink]

### Show Tags

1
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 well-shuffled cards. Find the probability that

$$\frac{13}{52}*\frac{12}{51}$$

(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.
_________________
Manager  Joined: 27 Apr 2008
Posts: 158
Re: Probability very good question set  [#permalink]

### Show Tags

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) = $$\frac{(first card non face + both cards face)}{52C2}$$
= $$\frac{40*12+12*11}{51*26}$$
= $$\frac{306}{663}$$
= $$\frac{102}{221}$$
Manager  Joined: 27 Apr 2008
Posts: 158
Re: Probability very good question set  [#permalink]

### Show Tags

Why would order matter? For the denominator you just want the # of combinations of getting 2 cards.
Math Expert V
Joined: 02 Sep 2009
Posts: 59095
Re: Probability very good question set  [#permalink]

### Show Tags

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.
_________________
Manager  Status: Its Wow or Never
Joined: 11 Dec 2009
Posts: 138
Location: India
Concentration: Technology, Strategy
GMAT 1: 670 Q47 V35 GMAT 2: 710 Q48 V40 WE: Information Technology (Computer Software)
Re: Probability very good question set  [#permalink]

### Show Tags

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.?
_________________
---------------------------------------------------------------------------------------
If you think you can,you can
If you think you can't,you are right.
Math Expert V
Joined: 02 Sep 2009
Posts: 59095
Re: Probability very good question set  [#permalink]

### Show Tags

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 non-face 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.
_________________
Manager  Status: Its Wow or Never
Joined: 11 Dec 2009
Posts: 138
Location: India
Concentration: Technology, Strategy
GMAT 1: 670 Q47 V35 GMAT 2: 710 Q48 V40 WE: Information Technology (Computer Software)
Re: Probability very good question set  [#permalink]

### Show Tags

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 non-face 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!!
_________________
---------------------------------------------------------------------------------------
If you think you can,you can
If you think you can't,you are right.
Manager  Joined: 27 Apr 2008
Posts: 158
Re: Probability very good question set  [#permalink]

### Show Tags

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.
Non-Human User Joined: 09 Sep 2013
Posts: 13602
Re: Two cards are drawn successively from a standard deck of 52  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: Two cards are drawn successively from a standard deck of 52   [#permalink] 28 Mar 2019, 04:46
Display posts from previous: Sort by

# Two cards are drawn successively from a standard deck of 52  