probabilty...set A & B..

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advita: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Mon Sep 13, 2010 8:06 pm; Thanked: 2 times

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.

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Source: Beat The GMAT — Problem Solving | Thu Jan 06, 2011 9:14 pm

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

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.

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towerSpider: 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

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

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

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RACHVIK: Senior | Next Rank: 100 Posts; Posts: 95; Joined: Wed Sep 22, 2010 8:26 amFollowed by: 1 member; Thanked: 1 times

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

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

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.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; 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.

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