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If an integer is to be randomly selected from set M above [#permalink]

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12 Jan 2008, 05:29

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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?

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

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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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

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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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.

Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]

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31 Mar 2012, 00:25

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 ?

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).

Re: M = {-6, -5, -4, -3, -2} T = {-2, -1, 0, 1, 2, 3} If an [#permalink]

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05 Oct 2012, 22:26

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

Re: If an integer is to be randomly selected from set M above [#permalink]

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04 Jul 2014, 11:03

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Re: If an integer is to be randomly selected from set M above [#permalink]

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17 Oct 2015, 09:10

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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

Total number of cases = 6*6 To get a negative result, we need to multiply numbers of negative signs. Set M has all the 6 negative numbers. Set T has 3 positive numbers, which when multiplied by negative numbers will yield a negative number

Hence we have 6*3 favourable cases.

Probability = 18/36 = 1/2 Option D
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Re: If an integer is to be randomly selected from set M above [#permalink]

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21 Oct 2016, 21:16

Hello from the GMAT Club BumpBot!

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Re: If an integer is to be randomly selected from set M above [#permalink]

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17 Dec 2016, 18:49

TeamGMATIFY wrote:

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

Total number of cases = 6*6 To get a negative result, we need to multiply numbers of negative signs. Set M has all the 6 negative numbers. Set T has 3 positive numbers, which when multiplied by negative numbers will yield a negative number

Re: If an integer is to be randomly selected from set M above [#permalink]

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14 Mar 2017, 20:07

kbulse wrote:

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

Set M has all negative numbers, so, the probability of selecting a negative number lies entirely on set M. Set M has 2 negative, 1 zero and 3 positive numbers. Thus, P(negative product) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Re: If an integer is to be randomly selected from set M above [#permalink]

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15 Mar 2017, 10:13

Shruti0805 wrote:

kbulse wrote:

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

Set M has all negative numbers, so, the probability of selecting a negative number lies entirely on set M. Set M has 2 negative, 1 zero and 3 positive numbers. Thus, P(negative product) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

This question would get much more tricky if 2/3 was one of the altenatives...

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 for the product of the two integers to be negative, one of them has to be negative and the other has to be positive. Since every integer in set M is negative, we must select a positive integer in set T. Thus, the probability of selecting a negative number in set M and then a positive number in set T is:

5/5 x 3/6 = 1/2

Answer: D
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Re: If an integer is to be randomly selected from set M above [#permalink]

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14 Apr 2017, 13:25

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

Any number picked from the first set will have the same properties (all -ve). A positive number is required from the second set to make the product -ve. 1/2 chance of picking a +ve number from the second set.

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