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# Q30: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2,

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Manager
Joined: 28 Jan 2006
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Q30: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, [#permalink]

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26 Aug 2006, 08:50
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Q30:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5
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All the best!!
shinewine

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Intern
Joined: 17 Apr 2006
Posts: 42

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26 Aug 2006, 10:49
shinewine wrote:
Q30:
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?

A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5

I go with D

Required probability = P(2 picks are of opposite sign)
We know that the first pick is always negative. Hence it is suffiencent to calculate probability that 2nd pick is positive. Since there are 3 nos. that are positive out of 6 in the set T
This is 3/6 or 1/2

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Current Student
Joined: 29 Jan 2005
Posts: 5201

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26 Aug 2006, 10:55
Probability = #favorable outcomes/#possible outcomes

neg*pos=neg: half of the numbers (3) in set T are pos.

probability = 3/6 ---> 1/2

(D)

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Senior Manager
Joined: 14 Jul 2005
Posts: 398

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26 Aug 2006, 11:19
Getting D....

Total no. of possible outcomes = 6*6 = 36

Total no. of outcomes with negative product = 6+6+6 (each for 1,2,3 members for set T)

Hence answer = 18/36 = 1/2

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Senior Manager
Joined: 14 Aug 2006
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26 Aug 2006, 16:44
D as well

For the outcome to be negative T has to be > 0 and positive

prob(T positive and >0) = = 3/6 = 1/2

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Manager
Joined: 20 Mar 2006
Posts: 200

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26 Aug 2006, 23:31
Total possibilities of selecting a number from M and one from T = 30
Possibilities with product is -ve = 15

Hence prob = 15/30 = 1/2

D

Heman

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Re: PS : Sets   [#permalink] 26 Aug 2006, 23:31
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