BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Probability and Permutations and Combinations

Expert replies
by bhumika.k.shah » Wed Jan 27, 2010 9:24 am
Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?
b) at least one of the cards is a picture card?
c) there is at most one heart?
d) each card has a different face value?

Detailed Explanations pleaseeee :-)
Join the discussion
Source: — Problem Solving |

by thephoenix » Wed Jan 27, 2010 9:32 am
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?

4C1*3C1*2C1/52C1*51C1*50C1=24/52*51*50
Join the discussion

by bhumika.k.shah » Wed Jan 27, 2010 9:36 am
?
thephoenix wrote:
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?

4C1*3C1*2C1/52C1*51C1*50C1=24/52*51*50

b) at least one of the cards is a picture card?
1-40C1/52C1=8/52


Detailed Explanations pleaseeee :-)
Join the discussion

by thephoenix » Wed Jan 27, 2010 9:40 am
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:


b) at least one of the cards is a picture card?

Detailed Explanations pleaseeee :-)
1-NONE IS PIC CARD
1-[(40C1/52C1)*(39C1/51C1)*(38C1/50C1)]
Last edited by thephoenix on Wed Jan 27, 2010 9:49 am, edited 1 time in total.
Join the discussion

by ajith » Wed Jan 27, 2010 9:41 am
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?
b) at least one of the cards is a picture card?
c) there is at most one heart?
d) each card has a different face value?

Detailed Explanations pleaseeee :-)
a) 4/52*3/51*2/50 (4 aces in a deck)

b) At least one Picture card can be subdivided into

1. Exactly 1 picture card 2. Exactly 2 Picture cards 3. Exactly 3 Picture cards

exactly one is picture + exactly two are picture + exactly 3 are picture
12/52*40/51*39/50+ 12/52*11/51*40/50+ 12/52*11/51*10/50


c) exactly 0 heart + exactly 1 heart
39/52*38/51*37/50 + 13/52*39/51*38/50

d) d) 1*48/51*44/50
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by thephoenix » Wed Jan 27, 2010 9:45 am
bhumika.k.shah wrote: FOR d
52C1*48*C1*44C1=52*48*44
Join the discussion

by ajith » Wed Jan 27, 2010 9:48 am
thephoenix wrote:
bhumika.k.shah wrote: FOR d
52C1*48*C1*44C1=52*48*44
The question asks the probability, ThePhoenix :)
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by thephoenix » Wed Jan 27, 2010 9:54 am
ajith wrote:
thephoenix wrote:
bhumika.k.shah wrote: FOR d
52C1*48*C1*44C1=52*48*44
The question asks the probability, ThePhoenix :)
YEAH THANKS
IT WILL BE 52/52 * 48/51 *44/50

THANKS FOR WAKING ME
Join the discussion

by bhumika.k.shah » Wed Jan 27, 2010 9:56 am
Is it the same as 4C3 / 52C3????
ajith wrote:
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?


a) 4/52*3/51*2/50 (4 aces in a deck)
Join the discussion

by ajith » Wed Jan 27, 2010 9:59 am
bhumika.k.shah wrote:Is it the same as 4C3 / 52C3????
ajith wrote:
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?


a) 4/52*3/51*2/50 (4 aces in a deck)

Yup it is the same as 4C3/52C3
Always borrow money from a pessimist, he doesn't expect to be paid back.
Join the discussion

by bhumika.k.shah » Wed Jan 27, 2010 7:01 pm
Hey ajith could you explain all of these in the PnC formula way???

Like i know for a.) the answer is 4C3/52C3

Thanks!
:)
ajith wrote:
bhumika.k.shah wrote:Three cards are drawn at random from a standard deck of playing cards
without replacement. What is the probability that:

a) all 3 cards are Aces?
b) at least one of the cards is a picture card?
c) there is at most one heart?
d) each card has a different face value?

Detailed Explanations pleaseeee :-)
a) 4/52*3/51*2/50 (4 aces in a deck)

b) At least one Picture card can be subdivided into

1. Exactly 1 picture card 2. Exactly 2 Picture cards 3. Exactly 3 Picture cards

exactly one is picture + exactly two are picture + exactly 3 are picture
12/52*40/51*39/50+ 12/52*11/51*40/50+ 12/52*11/51*10/50


c) exactly 0 heart + exactly 1 heart
39/52*38/51*37/50 + 13/52*39/51*38/50

d) d) 1*48/51*44/50
Join the discussion