probability easy (400-520)

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

probability easy (400-520)

by Night reader » Thu Dec 02, 2010 11:20 am

If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

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Source: Beat The GMAT — Problem Solving | Thu Dec 02, 2010 11:20 am

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pm; Location: Milpitas, CA; Thanked: 447 times; Followed by:88 members

by Rahul@gurome » Thu Dec 02, 2010 11:44 am

Night reader wrote:If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6

Total number of ways to select x and y = 3*4 = 12

Number of ways to select x and y such that xy is even,

If x = 1 or 3, y can be 4 or 6 --> (2 + 2) = 4 ways
If x = 2, y can be any of the four --> 4 ways

A total of 8 ways.

Required probability = 8/12 = 2/3

The correct answer is D.

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

by Night reader » Thu Dec 02, 2010 11:52 am

Rahul@gurome wrote:
Night reader wrote:If x is to be chosen at random from the set {1; 2; 3} and y is to be chosen at random from the set {4; 5; 6; 7}, what is the probability that xy will be even?

(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
Total number of ways to select x and y = 3*4 = 12

Number of ways to select x and y such that xy is even,
If x = 1 or 3, y can be 4 or 6 --> (2 + 2) = 4 ways
If x = 2, y can be any of the four --> 4 ways
A total of 8 ways.

Required probability = 8/12 = 2/3

The correct answer is D.

Thanks Rahul, that must be too easy for you.

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apex231: Master | Next Rank: 500 Posts; Posts: 127; Joined: Mon May 11, 2009 10:04 am; Thanked: 5 times

by apex231 » Sun Dec 05, 2010 3:56 am

I am wondering shouldn't it be 5/6?

P to select 2 = 1/3
P of even product = 1/3 * 1 = 1/3

P to select 4 = 1/4
P of even product = 1/4 * 1 = 1/4

P to select 6 = 1/4
P of even product = 1/4 * 1 = 1/4

Total P of even product = 1/3 + 1/4 + 1/4 = 5/6

Also, using combinations
when 2 is selected, no. of even products = 1 * 4 = 4
when 4 is is selected, no. of even products = 1 * 3 = 3
when 6 is is selected, no. of even products = 1 * 3 = 3

Total even products = 4+3+3 = 10
Total products = 3*4 = 12

Probability of even product = 10/12 = 5/6

Am i missing something?

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apex231: Master | Next Rank: 500 Posts; Posts: 127; Joined: Mon May 11, 2009 10:04 am; Thanked: 5 times

by apex231 » Sun Dec 05, 2010 3:58 am

apex231 wrote:I am wondering shouldn't it be 5/6?

P to select 2 = 1/3
P of even product = 1/3 * 1 = 1/3

P to select 4 = 1/4
P of even product = 1/4 * 1 = 1/4

P to select 6 = 1/4
P of even product = 1/4 * 1 = 1/4

Total P of even product = 1/3 + 1/4 + 1/4 = 5/6

Also, using combinations
when 2 is selected, no. of even products = 1 * 4 = 4
when 4 is is selected, no. of even products = 1 * 3 = 3
when 6 is is selected, no. of even products = 1 * 3 = 3

Total even products = 4+3+3 = 10
Total products = 3*4 = 12

Probability of even product = 10/12 = 5/6

Am i missing something?

Got it. I am counting combinations with 2 twice....