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If an integer is to be randomly selected from set M above

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If an integer is to be randomly selected from set M above [#permalink] New post 12 Jan 2008, 04:29
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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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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink] New post 31 Mar 2012, 00:34
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priyalr wrote:
Hello,

I got a doubt. Any no. mulitplied by 0 is 0 and 0 is neither +ve nor -ve. Going by this prop of 0, y shld we consider 0 as one of the nos. cause any no +ve or -ve multiplied by 0 is neither positive nor negative ?

Thnx.


It's not clear what you mean by "why should we consider 0 as one of the numbers"... Anyway:

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

In order the product of two multiples to be negative they must have different signs. Since Set M consists of only negative numbers then in order mt to be negative we should select positive number from set T, the probability of that event is 3/6=1/2, (since out of 6 number in the set 3 are positive).

Answer: D.

Hope it's clear.
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Re: good prob [#permalink] New post 12 Jan 2008, 04:40
marcodonzelli wrote:
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 think D is correct
in order to get negative integer either m should be negative and n positive
probability of m is negative =5/5=1
probability of n is positive 3/6=1/2
probability of m negative and m positive is 1*1/2=1/2
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Re: good prob [#permalink] New post 25 Aug 2008, 08:10
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marcodonzelli wrote:
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


P= No of ways the product of two integers is negative/All possible values
= 5*3/5*6 =1/2
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Re: good prob [#permalink] New post 27 Sep 2009, 05:53
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

Soln: Total number of possible ways of choosing two integers is = 5 * 6 = 30 ways
Now for the product of two integers to be chosen to be negative = they should be of opposite sign. Since set M has all negative numbers, thus we move to set T which has 3 positive numbers.
Thus total number of possible ways in which product will be negative is = 5 * 3 = 15

Probability that the product of the two integers will be negative
= 15/30
= 1/2
Ans is D
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Re: good prob [#permalink] New post 14 Feb 2010, 14:27
marcodonzelli wrote:
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


Negative prod pairs = 5 x 3 = 15
total pairs possible = 5c1 x 6c1 = 5 x 6 = 30

Probability = 15 / 30 = 1/2
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Re: good prob [#permalink] New post 14 Feb 2010, 23:00
If product has to be negative then m is -ve and n is +ve or m is +ve and n is -ve. But all elements of m are -ve hence we need n to be +ve.
0 has to be excluded as any number multiplied by 0 is 0 and it is neither +ve nor -ve.

p(m&n) = p(m)*p(n) = 1*3/6 = 1/2 hence D
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Re: good prob [#permalink] New post 14 Oct 2010, 17:40
All numbers in the first bank are negative so it has to select a positive from the second bank, but zero won't work since it would be zero.

5/5 * 3/6 = 15/30 => 1/2
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink] New post 30 Mar 2012, 23:25
Hello,

I got a doubt. Any no. mulitplied by 0 is 0 and 0 is neither +ve nor -ve. Going by this prop of 0, y shld we consider 0 as one of the nos. cause any no +ve or -ve multiplied by 0 is neither positive nor negative ?

Thnx.
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Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink] New post 05 Oct 2012, 21:26
Bunuel wrote:
priyalr wrote:
Hello,

I got a doubt. Any no. mulitplied by 0 is 0 and 0 is neither +ve nor -ve. Going by this prop of 0, y shld we consider 0 as one of the nos. cause any no +ve or -ve multiplied by 0 is neither positive nor negative ?

Thnx.


It's not clear what you mean by "why should we consider 0 as one of the numbers"... Anyway:

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

In order the product of two multiples to be negative they must have different signs. Since Set M consists of only negative numbers then in order mt to be negative we should select positive number from set T, the probability of that event is 3/6=1/2, (since out of 6 number in the set 3 are positive).

Answer: D.

Hope it's clear.


Why us 0 not considered as the product of any number will give non negetive
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Re: If an integer is to be randomly selected from set M above [#permalink] New post 04 Jul 2014, 10:03
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Re: If an integer is to be randomly selected from set M above   [#permalink] 04 Jul 2014, 10:03
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