need help with probability

Expert replies

abhishek07sep: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu Jul 12, 2012 11:10 am

by abhishek07sep » Wed Jul 18, 2012 9:19 am

1. from a pack of 52 cards 3 cards are drawn at random. what is the probability of getting an ace , a jack and a queen?

2. 4 cards are drawn at random from a deck of cards. find the probability of getting all the four cards of different numbers...

in need of complete explanation..

Like this post

Source: Beat The GMAT — Problem Solving | Wed Jul 18, 2012 9:19 am

kartikshah: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sun Jul 15, 2012 12:03 pmFollowed by: 1 member; Location: New York, USA; Thanked: 34 times

by kartikshah » Wed Jul 18, 2012 10:17 am

Solution to Question 01:

3 cards can be drawn from a pack of 52 cards in 52C3 ways:
= (52 * 51 * 50) / (3 *2 *1) ways

= 26 * 17 * 50 ways

Let A = event of getting an ace card, a jack card and a queen card
There are 4 ace cards, 4 jack cards and 4 queen cards

1 ace card can be drawn in 4C1 ways = 4 ways
1 jack card can be drawn in 4C1 ways = 4 ways
1 queen card can be drawn in 4C1 ways = 4 ways

Therefore, n(A) = 4 * 4 * 4 ways

Probability of this event = (4*4*4)/(26*17*50) = 16/5525

Like this post

confuse mind: Master | Next Rank: 500 Posts; Posts: 462; Joined: Wed Jan 19, 2011 1:08 pmFollowed by: 4 members; Thanked: 10 times

by confuse mind » Wed Jul 18, 2012 6:40 pm

Question - 2: 4 cards are drawn at random from a deck of cards. find the probability of getting all the four cards of different numbers...

Total number of way - 52C4
Favourable:
There are 10 number cards of each suite = 40

first card can be selected in 40C1 ways
second card can be selected in 36C1 ways
third - 32C1
forth - 28C1

4 cards can be selected in 40 * 36 * 32* 28 ways

But this has taken the order of selection into account
e.g.,

1 2 3 4
2 1 3 4

have been taken as 2 options in the above counting and thus divide by 4!

thus fovourable = (40 * 36 * 32* 28)/4!

divide favourable by total to get the answer

thanks

Like this post

abhishek07sep: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Thu Jul 12, 2012 11:10 am

by abhishek07sep » Wed Jul 18, 2012 8:07 pm

kartikshah wrote:Solution to Question 01:

3 cards can be drawn from a pack of 52 cards in 52C3 ways:
= (52 * 51 * 50) / (3 *2 *1) ways

= 26 * 17 * 50 ways

Let A = event of getting an ace card, a jack card and a queen card
There are 4 ace cards, 4 jack cards and 4 queen cards

1 ace card can be drawn in 4C1 ways = 4 ways
1 jack card can be drawn in 4C1 ways = 4 ways
1 queen card can be drawn in 4C1 ways = 4 ways

Therefore, n(A) = 4 * 4 * 4 ways ace j, a

Probability of this event = (4*4*4)/(26*17*50) = 16/5525

if for selecting an ace, a jack and a king i do,
12c1*8c1*4c1 what answer will this give to me?

Like this post

eagleeye: Legendary Member; Posts: 520; Joined: Sat Apr 28, 2012 9:12 pmFollowed by: 49 members; Thanked: 339 times; GMAT Score:770

by eagleeye » Wed Jul 18, 2012 9:49 pm

abhishek07sep wrote:
if for selecting an ace, a jack and a king i do,
12c1*8c1*4c1 what answer will this give to me?

If you go this route, you can still get the right answer, as long as you do the division the correct way.
Here are both the questions solved using this:

1st question:
Required probability = (12C1*8C1*4C1)/(52C1*51C1*50C1) = (12*8*4)/(52*51*50) = 16/5525

2nd question: In the same way as above,
Required probability = (40*36*32*28)/(52*51*50*49)

Let me know if this helps

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review