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

Redeem

probabilty...set A & B..

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 62
Joined: Mon Sep 13, 2010 8:06 pm
Thanked: 2 times

probabilty...set A & B..

by advita » Thu Jan 06, 2011 9:14 pm
#Q1:
A={2,4,4,6,8}
B={2,3,4,6,6,8,12,12}

two integers are chosen from above sets, 1 from A and 1 from B. what is the probability that the product of two integer is 24.

1st- no options..

2nd- my query: i know total possibilities=5*8=40 ( 5 elements for A ,7 for B)
now.... numerator will be what...??

method 1:
(2,12) (2,12) ........(for one 2 from A and two 12 from B)
(4,6) (4,6) ......... ( for 1 st 4 from A and 1st 6 from B)
(4, 6) (4,6) .............( for 2nd 4 and 6 from A and B)
(6,4) and (8,3).

so total =8.

or

method 2:

(2,12) (4,6) (8,3) ..... only one elements from each...(if they are in multiple numbers..)

or.....what......

( a little varied form of question that i got from net...)[/i]

please clarify...thanks...!!!
Last edited by advita on Thu Jan 06, 2011 11:59 pm, edited 1 time in total.
Top
Source: — Problem Solving |
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Thu Jan 06, 2011 9:43 pm
advita wrote:#Q1:
A={2,4,4,6,8}
B={2,3,4,6,6,8,12,12}

two integers are chosen from above sets, 1 from A and 1 from B. what is the probability that the product of two integer is 24.
A={2^1,2^2,2^2,2^1 * 3^1,2^3}
B={2^1,3^1,2^2,2^1 * 3^1,2^1 * 3^1,2^3,2^2 * 3^1,2^2 * 3^1}

24=2^3 * 3^1
P(A*B=24) OR P(2^3 * 3^1):
P(2^1 `A)&P(2^2 * 3^1 `B)=1/5 * 2/8 = 1/20
P(2^2 `A)&P(2^1 * 3^1 `B)=2/5 * 2/8 = 1/10
P(2^1 * 3^1 `A)&P(2^2 `B)=1/5 * 1/8 = 1/40
P(2^3 `A)&P(3^1 `B)= 1/5 * 1/8 = 1/40


P(2^3 * 3^1)= 1/20 + 1/10 + 1/40 + 1/40 = 8/40 OR 1/5
Last edited by Night reader on Fri Jan 07, 2011 1:49 am, edited 1 time in total.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 155
Joined: Mon Dec 13, 2010 11:02 am
Thanked: 3 times

by towerSpider » Thu Jan 06, 2011 11:34 pm
advita wrote:#Q1:
A={2,4,4,6,8}
B={2,3,4,6,6,8,12,12}

two integers are chosen from above sets, 1 from A and 1 from B. what is the probability that the product of two integer is 24.

1st- no options..

2nd- my query: i know total possibilities=5*8=40 ( 5 elements for A ,7 for B)
now.... numerator will be what...??

2,12 2,12 (for 2 form a and two 12 from B) 4,6 4,6 ( for 1 st 4 from A and 1st 6 from B) 4, 6, 4,6 ( for 2nd 4 and 6 from A and B) 8,3.

or 2,12 4,6 8,3... only one elements from each...(if they are in multiple numbers..)

or ,,,what......

( a little varied form of question that i got from net...)

please clarify...thanks...!!!
Total possibilities: 40
Favoured: 8
Answer: 1/5
Top
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Fri Jan 07, 2011 2:11 am
advita wrote:#Q1:
A={2,4,4,6,8}
B={2,3,4,6,6,8,12,12}

two integers are chosen from above sets, 1 from A and 1 from B. what is the probability that the product of two integer is 24.
the less time consuming approach follows

2 integers (selections) to pair and make the product for 24 from set A={2,4,4,6,8} and set B={2,3,4,6,6,8,12,12} are

2-12 OR 1 two & 2 twelves => 1*2
4-6 OR 1 four & 2 sixes => 1*2
4-6 OR 1 four & 2 sixes => 1*2
6-4 OR 1 six & 1 four => 1*1
8-3 OR 1 eight & 1 three => 1*1

total 8 selections; counting total ways of possible selections of two integers from set A and B => 5*8=40

probability = the number of favorable outcomes/total outcomes => 8/40 OR 1/5
Top
Senior | Next Rank: 100 Posts
Posts: 95
Joined: Wed Sep 22, 2010 8:26 am
Thanked: 1 times
Followed by:1 members

by RACHVIK » Fri Jan 07, 2011 2:33 am
Whats the OA. Why not 1/6??

Since digits are repeated in both sets, sets with common elements are repeated. The total possibilities work out to 24 and 4 favorable outcomes.

Please confirm.

Thanks
Rachvik
Top
Legendary Member
Posts: 1337
Joined: Sat Dec 27, 2008 6:29 pm
Thanked: 127 times
Followed by:10 members

by Night reader » Fri Jan 07, 2011 2:47 am
RACHVIK wrote:Whats the OA. Why not 1/6??

Since digits are repeated in both sets, sets with common elements are repeated. The total possibilities work out to 24 and 4 favorable outcomes.

Please confirm.

Thanks
good question, that's why I made myself to solve for this problem by using individual probabilities first. If you noticed the integer 4 is repeated two times in set A; the probability of selecting the integer 4 from set A is also 2/5.
hope this clarifies your doubt about repeating values.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Fri Jan 07, 2011 3:31 am
advita wrote:A={2,4,4,6,8}
B={2,3,4,6,6,8,12,12}

Two integers are chosen from above sets, 1 from A and 1 from B. what is the probability that the product of two integer is 24.
Total number of possible selections of two integers such that one is from set A and the other from set B = (Number of possible selections of 1 element out of 5)*(Number of possible selections of 1 element out of 8) = 5*8 = 40

Now possible selections to get a product of 24:
  • 1. 2 from set A and 12 from set B ---> 2 possibility
    2. 4 from set A and 6 from set B ---> 4 possibility
    3. 6 from set A and 4 from set B ---> 1 possibility
    4. 8 from set A and 3 from set B ---> 1 possibility
A total of 8 possibilities.

Hence, the required probability = 8/40 = 1/5

Also we can calculate the individual probability of selecting the pairs (2, 12), (3, 8), and (4, 6) and add them to calculate the probability.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Post new topic Post Reply
• Page 1 of 1